Limit and Continuity(I) | NEB Class 11 Mathematics Notes

Chapter Overview

Calculus begins here. A limit asks: what value does f(x) approach as x gets arbitrarily close to some number a? — without x actually reaching a. This seemingly simple question unlocks the entire machinery of derivatives, integrals, and beyond. This chapter is the gatekeeper of Class 11–12 Mathematics: you cannot understand derivatives or antiderivatives without first mastering how limits resolve 'illegal' expressions like 0/0. Continuity then asks: is there a break in the graph at a point?

What this chapter covers

Intuitive and Formal Definition of a LimitProperties of LimitsThe 7 Indeterminate FormsAlgebraic Limit Theorem: (xⁿ − aⁿ)/(x−a)Trigonometric Limits: sin θ/θ and tan θ/θExponential Limits: (eˣ−1)/x and (aˣ−1)/xLogarithmic Limits: ln(1+x)/xLimits at Infinity and Infinite LimitsLeft-Hand and Right-Hand LimitsDefinition and Conditions for ContinuityTypes of Discontinuity

Learning Objectives

1

Define a limit intuitively and explain why we approach but don't reach the point

2

Identify the 7 indeterminate forms and explain why they cannot be resolved by simple arithmetic

3

Evaluate limits using: direct substitution, factorisation, rationalisation, and dividing by highest power

4

Apply the standard trigonometric limits (sin θ/θ = 1, tan θ/θ = 1) by matching the angle inside and outside

5

Apply exponential and logarithmic limit formulas to evaluate complex expressions

6

Use the algebraic limit theorem (xⁿ−aⁿ)/(x−a) = naⁿ⁻¹ including for fractional n

7

Prove the fundamental trig limit using the Squeeze Theorem with triangle and sector areas

8

Check the three conditions of continuity at a point and classify types of discontinuity

The One Idea Behind This Chapter

Key Insight

A limit does NOT ask what happens at x = a — it asks what the function is heading towards as x approaches a from both sides.

This is why we can evaluate $\lim_{x \to 2} \frac{x^2-4}{x-2}$ even though substituting x=2 gives 0/0. The function is undefined at x=2 but still has a perfectly well-defined limit of 4 there. Limits describe the journey, not the destination.

Golden rule: if direct substitution gives a valid number (not 0/0, ∞/∞, etc.), that IS the limit. Only do algebra when substitution fails.

Key Definitions

Limit

A function $f(x)$ tends to a limit $l$ as $x$ approaches $a$ if the value of $f(x)$ can be made as close as we please to $l$ by taking $x$ sufficiently close to $a$ (but $x \neq a$).

$$\lim_{x \to a} f(x) = l$$

Left-Hand Limit (LHL)

The limit of $f(x)$ as $x$ approaches $a$ from the left side (from values smaller than $a$). Written as $x \to a^-$.

$$\lim_{x \to a^-} f(x)$$

Right-Hand Limit (RHL)

The limit of $f(x)$ as $x$ approaches $a$ from the right side (from values larger than $a$). Written as $x \to a^+$.

$$\lim_{x \to a^+} f(x)$$

Existence of Limit

A limit $\lim_{x \to a} f(x)$ exists if and only if the Left-Hand Limit and Right-Hand Limit are equal. If they differ, the limit does not exist.

$$\lim_{x \to a} f(x) \text{ exists} \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$

Indeterminate Forms

Expressions that cannot be directly evaluated by simple arithmetic — they require further algebraic manipulation. There are 7 indeterminate forms.

$$\frac{0}{0},\; \frac{\infty}{\infty},\; \infty - \infty,\; 0 \cdot \infty,\; 0^0,\; 1^\infty,\; \infty^0$$

Continuity at a Point

A function $f$ is continuous at $x = a$ if all three conditions hold: (1) $f(a)$ exists, (2) $\lim_{x \to a} f(x)$ exists, (3) $\lim_{x \to a} f(x) = f(a)$. If any condition fails, $f$ has a discontinuity at $x = a$.

$$\lim_{x \to a} f(x) = f(a)$$

Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ near $x = a$, and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = l$, then $\lim_{x \to a} f(x) = l$. Used to prove the trig limit $\sin\theta/\theta \to 1$.

The 7 Indeterminate Forms — Why They Exist

These 7 forms arise when direct substitution gives an expression whose value is mathematically ambiguous — different problems involving these forms can give completely different answers. They signal that you must do more algebra before the limit can be determined.

$$\frac{0}{0},\; \frac{\infty}{\infty},\; \infty - \infty,\; 0 \cdot \infty,\; 0^0,\; 1^\infty,\; \infty^0$$

Why it works (Derivation):

10/0 — the most common: $\frac{x^2-4}{x-2}$ gives 0/0 at x=2, but the actual limit is 4. Another 0/0 expression: $\frac{x^2}{x}$ at x=0 gives 0. The form 0/0 can equal ANY finite number or infinity depending on the problem.

2∞/∞: $\frac{3x^2+1}{2x^2}$ gives ∞/∞ at x→∞, but the answer is 3/2. Yet $\frac{x}{x^3}$ also gives ∞/∞ but the answer is 0. Same form, different answers.

3∞−∞: $\sqrt{x^2+x} - x$ gives ∞−∞, yet equals 1/2. $\sqrt{x} - x$ gives ∞−∞ but equals −∞. Again, same form, wildly different answers — that's why it's indeterminate.

4NOT indeterminate: $0^\infty = 0$. This one always gives 0 (something infinitely small raised to a huge power shrinks to zero). So $0^\infty$ is NOT in the indeterminate list.

Example: Test: is $\frac{0}{0}$ equal to 1? Try: $\frac{x-x}{x-x}$ — factoring gives $\frac{0}{0}$ but the expression always equals 1. Now try $\frac{x^2-x}{x-1}$ at x=1 — also $\frac{0}{0}$ but limit is $1^2 - 1 = 0$... actually $\frac{x(x-1)}{x-1} = x \to 1$. So 0/0 can equal 1 here. The form alone tells you nothing — do the algebra!

When to use: Whenever direct substitution gives one of the 7 forms, stop and apply the appropriate technique (factor, rationalize, use standard formula).

Left-Hand Limit vs Right-Hand Limit

The limit exists only when both sides agree. This is the fundamental test for limit existence at a point.

Left-Hand Limit

$$\lim_{x \to a^-} f(x)$$

x approaches a from the left (x < a)

Written as $x \to a^-$ (a superscript minus)

Substitute $x = a - h$ where $h \to 0^+$

Represents function behaviour just to the left of the point

Can be different from the right-hand limit for piecewise or modulus functions

VS

Right-Hand Limit

$$\lim_{x \to a^+} f(x)$$

x approaches a from the right (x > a)

Written as $x \to a^+$ (a superscript plus)

Substitute $x = a + h$ where $h \to 0^+$

Represents function behaviour just to the right of the point

Used for functions with different definitions on different sides

Limit exists ⟺ LHL = RHL. If LHL ≠ RHL, write 'limit does not exist'. Most NEB questions on this involve piecewise functions or |x| expressions.

All Standard Limit Formulas with Derivations

Constant Rule

\lim_{x \to a} k = k

The limit of a constant is always the constant, regardless of what x approaches.

1A constant function $f(x) = k$ has the same value k for every single value of x — no matter how close x gets to a.

2Since f(x) = k always, it is trivially true that f(x) can be made as close as we please to k. No computation needed.

3Therefore $\lim_{x \to a} k = k$ for any value of a.

Algebraic Limit Theorem

\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}

Works for all rational n, not just positive integers. The most versatile formula in this chapter.

1For positive integer n: Use the algebraic identity: $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1})$. This identity has exactly n terms in the second factor.

2Substitute into the limit: $\lim_{x \to a} \frac{(x-a)(x^{n-1} + x^{n-2}a + \cdots + a^{n-1})}{x-a}$. Since $x \to a$ means $x \neq a$, the factor $(x-a)$ is never zero, so we can cancel it.

3After cancellation, substitute $x = a$: each of the n terms becomes $a^{n-1}$. Summing n copies: $$\lim_{x \to a} \frac{x^n - a^n}{x-a} = \underbrace{a^{n-1} + a^{n-1} + \cdots + a^{n-1}}_{n\text{ terms}} = na^{n-1}$$

Fundamental Trigonometric Limit

\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \quad (\theta \text{ in radians})

θ must be in radians. This limit is the backbone of all trig limit problems.

1Geometric setup: Consider a unit circle (radius 1). For a small positive angle θ (in radians), compare three areas: Area of △OPL < Area of sector OPA < Area of △OQA. Computing: $\frac{1}{2}\sin\theta\cos\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$.

2Apply Squeeze Theorem: Divide throughout by $\frac{1}{2}\sin\theta$ (positive for small θ > 0): $\cos\theta < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$. Taking reciprocals (which reverses inequalities): $\cos\theta < \frac{\sin\theta}{\theta} < \frac{1}{\cos\theta}$.

3Squeeze: As $\theta \to 0$, $\cos\theta \to 1$ and $\frac{1}{\cos\theta} \to 1$. So $\frac{\sin\theta}{\theta}$ is squeezed between two quantities both approaching 1. By the Squeeze Theorem: $$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$$

Tangent Limit

\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1

Follows directly from the sine limit combined with cos θ → 1.

1Write $\tan\theta = \frac{\sin\theta}{\cos\theta}$.

2Split the fraction: $\frac{\tan\theta}{\theta} = \frac{\sin\theta}{\theta} \cdot \frac{1}{\cos\theta}$.

3Take the limit: $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} \cdot \lim_{\theta \to 0} \frac{1}{\cos\theta} = 1 \cdot \frac{1}{1} = 1$.

Cosine Limit

\lim_{\theta \to 0} \cos \theta = 1

Cosine is continuous at 0, so direct substitution works.

1Cosine is a continuous function everywhere. For continuous functions, the limit equals the function value.

2$\lim_{\theta \to 0} \cos\theta = \cos(0) = 1$.

3Alternative derivation: use $\cos\theta = 1 - 2\sin^2(\theta/2)$. As $\theta \to 0$, $\sin(\theta/2) \to 0$, so $\cos\theta \to 1 - 0 = 1$.

Exponential Limit (base e)

\lim_{x \to 0} \frac{e^x - 1}{x} = 1

This is the definition of the derivative of eˣ at x=0. Direction: never substitute directly — always match the pattern ax/(ax).

1Using the Taylor series expansion: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

2So $e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

3Dividing by $x$: $\frac{e^x-1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots$. As $x \to 0$, all terms with x vanish: $$\lim_{x \to 0} \frac{e^x-1}{x} = 1$$

Exponential Limit (base a)

\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

Generalisation of the base-e formula. When a = e, ln e = 1, giving the previous formula.

1Write $a^x = e^{x \ln a}$, so $a^x - 1 = e^{x \ln a} - 1$.

2Multiply and divide by $\ln a$: $\frac{a^x-1}{x} = \frac{e^{x\ln a}-1}{x\ln a} \cdot \ln a$.

3Let $t = x\ln a$. As $x \to 0$, $t \to 0$. So: $\frac{e^t - 1}{t} \cdot \ln a \to 1 \cdot \ln a = \ln a$.

Logarithmic Limit

\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1

Natural log version. For log base a: divide by ln a.

1Using Taylor series: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ for $|x| < 1$.

2Divide by $x$: $\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots$.

3As $x \to 0$: all terms with x vanish, leaving: $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$.

1 − cos θ Identity (Key for Hard Problems)

1 - \cos\theta = 2\sin^2\!\left(\frac{\theta}{2}\right)

Not a formula for a limit, but a crucial identity used inside limits. Turns every (1−cos) problem into a squared sine limit.

1From the double-angle identity: $\cos 2\alpha = 1 - 2\sin^2\alpha$.

2Set $\alpha = \theta/2$: $\cos\theta = 1 - 2\sin^2(\theta/2)$.

3Rearranging: $1 - \cos\theta = 2\sin^2(\theta/2)$. So $\lim_{x \to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{2\sin^2(x/2)}{x^2} = \frac{2}{4}\left(\frac{\sin(x/2)}{x/2}\right)^2 = \frac{1}{2}$.

Continuity — Three Conditions

A function $f$ is continuous at $x = a$ if there is no break, jump, or hole in its graph at that point. Formally, three conditions must ALL hold simultaneously.

$$\text{Continuous at }a \iff \lim_{x \to a} f(x) = f(a)$$

Why it works (Derivation):

1Condition 1 — f(a) exists: The function must be defined at x = a. If there is a hole or the function is undefined there (like f(x) = 1/x at x=0), continuity fails immediately.

2Condition 2 — lim f(x) exists: The Left-Hand Limit and Right-Hand Limit must be equal. If the function jumps (like the floor function at integers), the limit doesn't exist and continuity fails.

3Condition 3 — limit equals function value: Even if both the limit and f(a) exist, they must be equal. A 'removable discontinuity' (hole) occurs when the limit exists but f(a) is defined differently or not at all at that point.

4Types of discontinuity: (a) Removable — limit exists but ≠ f(a) or f(a) undefined. Fix: redefine f(a). (b) Jump — LHL ≠ RHL. Cannot be fixed. (c) Infinite — limit is ±∞. Cannot be fixed. (d) Oscillatory — limit doesn't exist due to oscillation (e.g., sin(1/x) at x=0).

Example: Check continuity of $f(x) = \frac{x^2-4}{x-2}$ at $x=2$: Condition 1 FAILS (f(2) is undefined). It has a removable discontinuity at x=2. We can remove it by defining f(2)=4.

When to use: In NEB exams: either check if a function is continuous at a point (verify all 3 conditions) OR find the value of a constant that makes a piecewise function continuous.

Which Technique to Use? — Decision Tree

Every limit problem falls into one of these categories. Check in order.

1

Step 1 — Try Direct Substitution first. Substitute x = a directly. If you get a real number (not one of the 7 indeterminate forms), that IS the answer. You're done.

Did direct substitution give 0/0 or ∞/∞?

Yes

Continue to next step — you need algebra

No

If result is a real number → that's the limit. If result is ∞/0 or −∞ → limit is ±∞ or DNE

Does the expression involve polynomials or (xⁿ − aⁿ) type?

Yes

Factorise and cancel the common factor (x−a). Use the algebraic limit theorem for xⁿ−aⁿ form.

No

Continue ↓

Does the expression involve square roots (√x − √a)?

Yes

Rationalise by multiplying by the conjugate (√x + √a)/(√x + √a). This removes the root from the numerator.

No

Continue ↓

Does the expression involve sin, cos, or tan?

Yes

Use mirror trick: make sin(kx)/kx = 1. Use 1−cosx = 2sin²(x/2) to convert cosine problems to sine.

No

Continue ↓

Is x → ∞ (or x → −∞)?

Yes

Divide every term by the highest power of x present. Terms with x in denominator → 0.

No

Use exponential/log formulas: (eˣ−1)/x = 1, (aˣ−1)/x = ln a, ln(1+x)/x = 1

Memory Tricks — Never Miss These in Exams

Trick: Four quick rules that save minutes in NEB exams.

Steps to Remember:

The Mirror Trick for trig: The angle inside $\sin$ and the denominator must be mirrors (identical). Have $\sin 5x$? You need $5x$ below. Multiply and divide by 5: $\frac{\sin 5x}{x} = 5 \cdot \frac{\sin 5x}{5x} \to 5 \cdot 1 = 5$.
Factor or Rationalise for 0/0: Polynomials → factorise. Square roots → rationalise with conjugate. Never panic at 0/0 — it always has a hidden value waiting to emerge.
Degrees to Radians: $\lim_{x \to 0} \frac{\sin x°}{x}$ is NOT 1. Convert first: $x° = \frac{\pi x}{180}$ radians. So the limit becomes $\frac{\pi}{180}$.
∞ problems — divide by highest power: For $\lim_{x \to \infty} \frac{P(x)}{Q(x)}$: if degree of P = degree of Q → ratio of leading coefficients. If degree of P < degree of Q → 0. If degree of P > degree of Q → ±∞.
1−cos shortcut: Whenever you see $1 - \cos x$, immediately write $= 2\sin^2(x/2)$. This converts it to a sine limit which you know how to handle. Result: $\lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}$.

Mnemonic: Mirror trig | Factor/Rationalise 0/0 | Degrees→Radians | Divide by highest power | 1−cosx = 2sin²(x/2)

Common Mistakes

These mistakes cost the most marks in NEB exams for this chapter.

Ignoring Radians: $\lim_{x \to 0} \frac{\sin x°}{x}$ is $\frac{\pi}{180}$, NOT 1. The formula $\frac{\sin\theta}{\theta} = 1$ requires θ in radians. The geometric proof breaks down for degrees.
Premature substitution — dropping 'lim': Always keep writing $\lim_{x \to a}$ until the final step where you actually substitute. Dropping it early is a notation error that loses marks.
Treating 0/0 as zero: 0/0 is an indeterminate form, not zero. Students write $\frac{0}{0} = 0$ and stop. You must factor, rationalise, or use a formula to find the true limit.
Dividing by a trig function and losing solutions: When solving continuity problems involving piecewise functions, always check LHL = RHL = f(a) separately.
Forgetting the three conditions for continuity: Students check only the limit but forget to verify that f(a) is defined. A function can have a limit at a point but still be discontinuous there.
Wrong degree comparison for ∞: $\frac{3x^3+2x}{5x^2+1}$ as $x\to\infty$ — numerator is degree 3, denominator degree 2, so answer is $\infty$ (not 3/5). Only use the coefficient ratio when degrees are equal.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners test these edge cases consistently. Know them cold.

The geometric proof of $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$ using areas of triangles and sectors is the most asked long question in NEB. Draw the unit circle, show all three areas, apply Squeeze Theorem.
$0^\infty$ is NOT an indeterminate form — it equals 0. Students often include it in the list of 7. The 7 are: $\frac{0}{0}, \frac{\infty}{\infty}, \infty-\infty, 0\cdot\infty, 0^0, 1^\infty, \infty^0$.
For $\lim_{x\to a}\frac{x^{1/3}-a^{1/3}}{x^{1/2}-a^{1/2}}$: divide both by $(x-a)$, use algebraic theorem separately on top and bottom, then divide the results.
Problems of the form $\lim_{x\to 0}\frac{e^{5x}-e^{2x}}{x}$ — split as $(e^{5x}-1)/x - (e^{2x}-1)/x = 5(1) - 2(1) = 3$. Don't try L'Hôpital's rule.
Continuity at endpoints: for a closed interval $[a,b]$, continuity at $a$ requires only the RHL to equal f(a), and at $b$ only the LHL.
If asked to find k for continuity in a piecewise function: set up LHL = RHL = f(at junction), solve for k. Don't just set one side equal to the function value.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Direct substitution gives a real number	That's the answer — no work needed
0/0 with polynomial expressions	Factorise numerator and/or denominator, cancel (x−a)
0/0 with square roots (√x − √a)	Rationalise by multiplying by the conjugate
xⁿ − aⁿ form divided by (x−a)	Algebraic theorem: naⁿ⁻¹
sin(kx)/x or tan(kx)/x as x→0	Mirror trick: multiply-divide by k, apply sin(kx)/(kx)→1
1 − cos x anywhere in numerator	Replace with 2sin²(x/2), then use sin formula
x → ∞, polynomial/polynomial	Divide all terms by highest power of x
x → ∞, √(x²+...)	Factor out x from inside root, or rationalise
(eˣ − 1)/x or (aˣ − 1)/x as x→0	Standard formulas: 1 and ln a respectively
ln(1+x)/x as x→0	Standard formula: 1. For kx, multiply-divide by k
LHL and RHL check needed	Piecewise functions, \|x\|, floor/ceil functions
Check continuity at a point	Verify all 3 conditions: f(a) exists, limit exists, limit = f(a)
Find k for continuity	Set LHL = RHL = f(junction point), solve for k

Solved Examples (Easy → Hard)

1

Evaluate $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.

Easy

Solution

Step 1 — Try direct substitution: x=3 gives (9−9)/(3−3) = 0/0. Indeterminate — must factorise.

Step 2 — Factorise:
$$\frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3}$$

Step 3 — Cancel (x−3): Since $x \to 3$ means $x \neq 3$, we can divide:
$$= \lim_{x \to 3} (x+3) = 3 + 3 = 6$$

∴ 6

2

Evaluate $\lim_{x \to 0} \frac{\sin 7x}{x}$.

Easy

Solution

Mirror trick: Need $\frac{\sin 7x}{7x}$ pattern. Multiply and divide by 7:

$$\lim_{x \to 0} \frac{\sin 7x}{x} = 7 \cdot \lim_{x \to 0} \frac{\sin 7x}{7x}$$

As $x \to 0$, also $7x \to 0$. Apply $\frac{\sin t}{t} \to 1$ with $t = 7x$:

$$= 7 \times 1 = 7$$

∴ 7

3

Evaluate $\lim_{x \to \infty} \frac{3x^2+1}{2x^2-5}$.

Medium

Solution

x → ∞: divide all terms by highest power (which is x²):

$$= \lim_{x \to \infty} \frac{\frac{3x^2}{x^2}+\frac{1}{x^2}}{\frac{2x^2}{x^2}-\frac{5}{x^2}} = \lim_{x \to \infty} \frac{3 + \frac{1}{x^2}}{2 - \frac{5}{x^2}}$$

As $x \to \infty$: $\frac{1}{x^2} \to 0$ and $\frac{5}{x^2} \to 0$:

$$= \frac{3+0}{2-0} = \frac{3}{2}$$

∴ 3/2

4

Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Medium

Solution

Use the identity $1-\cos x = 2\sin^2\!\left(\frac{x}{2}\right)$:

$$\lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2}$$

Rewrite $x^2 = 4 \cdot (x/2)^2$:

$$= \frac{2}{4} \cdot \lim_{x \to 0} \left(\frac{\sin(x/2)}{x/2}\right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}$$

∴ 1/2

5

Evaluate $\lim_{x \to \infty} (\sqrt{x^2+x} - x)$.

Hard

Solution

Direct approach gives ∞−∞ (indeterminate). Rationalise by multiplying by conjugate:

$$= \lim_{x \to \infty} \frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x} = \lim_{x \to \infty} \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}$$

$$= \lim_{x \to \infty} \frac{x}{\sqrt{x^2+x}+x}$$

Divide by x (for x > 0, $\frac{1}{x} = \frac{1}{\sqrt{x^2}}$):

$$= \lim_{x \to \infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{\sqrt{1+0}+1} = \frac{1}{2}$$

∴ 1/2

MCQ Practice

Mixed Level — MCQ

1

What is the value of $\lim_{x \to 5} 7$?

Correct!

Incorrect. Correct: 7

Solution

For a constant function $f(x) = 7$, the output is always 7 regardless of x.

$\lim_{x \to 5} 7 = 7$

No computation needed — limit of a constant = the constant.

∴ 7

2

Which of these is NOT an indeterminate form?

Correct!

Incorrect. Correct: $0^\infty$

Solution

The 7 indeterminate forms are: $\frac{0}{0},\; \frac{\infty}{\infty},\; \infty-\infty,\; 0\cdot\infty,\; 0^0,\; 1^\infty,\; \infty^0$.

$0^\infty$ is NOT in this list because it always evaluates to 0 — no ambiguity.

∴ $0^\infty$

3

The value of $\lim_{x \to 0} \frac{\sin 2x}{x}$ is:

Correct!

Incorrect. Correct: 2

Solution

Multiply and divide by 2:
$$\lim_{x \to 0} \frac{\sin 2x}{x} = 2 \cdot \lim_{x \to 0} \frac{\sin 2x}{2x} = 2 \times 1 = 2$$

(As $x \to 0$, also $2x \to 0$, so $\frac{\sin 2x}{2x} \to 1$.)

∴ 2

4

$\lim_{x \to \infty} \frac{1}{x}$ equals:

Correct!

Incorrect. Correct: 0

Solution

As the denominator $x$ grows larger and larger without bound, the fraction $\frac{1}{x}$ gets smaller and smaller, approaching 0.

$$\lim_{x \to \infty} \frac{1}{x} = 0$$

∴ 0

5

The value of $\lim_{x \to a} \frac{x^3 - a^3}{x - a}$ is:

Correct!

Incorrect. Correct: $3a^2$

Solution

Using the algebraic limit theorem $\lim_{x \to a}\frac{x^n-a^n}{x-a} = na^{n-1}$:

Here $n = 3$:
$$= 3 \cdot a^{3-1} = 3a^2$$

∴ $3a^2$

6

$\lim_{x \to 0} \frac{e^x - 1}{x}$ is:

Correct!

Incorrect. Correct: 1

Solution

Standard exponential limit formula:
$$\lim_{x \to 0} \frac{e^x-1}{x} = 1$$

Derivation: $e^x = 1 + x + x^2/2! + \cdots$, so $\frac{e^x-1}{x} = 1 + x/2! + \cdots \to 1$.

∴ 1

7

Evaluate $\lim_{x \to 0} \frac{\log(1+x)}{x}$:

Correct!

Incorrect. Correct: 1

Solution

Standard logarithmic limit formula:
$$\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$$

This also holds when 'log' means natural log. For log base a: result would be $\frac{1}{\ln a}$.

∴ 1

8

$\lim_{x \to 0} \cos x$ is:

Correct!

Incorrect. Correct: 1

Solution

Cosine is a continuous function at $x = 0$, so we can directly substitute:
$$\lim_{x \to 0} \cos x = \cos(0) = 1$$

∴ 1

9

In the formula $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, $\theta$ must be in:

Correct!

Incorrect. Correct: Radians

Solution

The formula $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$ is derived using the arc length formula $s = r\theta$, which is valid only when $\theta$ is measured in radians.

In degrees: $\lim_{x \to 0} \frac{\sin x°}{x} = \frac{\pi}{180} \neq 1$.

∴ Radians

10

$\lim_{x \to 1} \frac{x^{10}-1}{x-1}$ equals:

Correct!

Incorrect. Correct: 10

Solution

Using the algebraic limit theorem with $n = 10$ and $a = 1$:
$$\lim_{x \to 1} \frac{x^{10}-1}{x-1} = 10 \cdot 1^{10-1} = 10 \cdot 1^9 = 10$$

∴ 10

Short Questions (2 Marks Each)

1

Evaluate $\lim_{x \to 2} \frac{x^2-5x+6}{x-2}$.

[2 marks]

Solution

Direct substitution gives 0/0. Factorise:

$x^2-5x+6 = (x-2)(x-3)$

$$\lim_{x \to 2} \frac{(x-2)(x-3)}{x-2} = \lim_{x \to 2}(x-3) = 2-3 = -1$$

∴ −1

2

Evaluate $\lim_{x \to 0} \frac{\sqrt{1+x}-1}{x}$.

[2 marks]

Solution

0/0 — rationalise by multiplying by conjugate $(\sqrt{1+x}+1)$:

$$= \lim_{x \to 0} \frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)}$$

$$= \lim_{x \to 0} \frac{x}{x(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{1+x}+1} = \frac{1}{1+1} = \frac{1}{2}$$

∴ 1/2

3

Evaluate $\lim_{x \to a} \frac{x^5-a^5}{x-a}$.

[2 marks]

Solution

Apply the algebraic limit theorem $\lim_{x \to a}\frac{x^n-a^n}{x-a} = na^{n-1}$ with $n = 5$:

$$= 5 \cdot a^{5-1} = 5a^4$$

∴ $5a^4$

4

Evaluate $\lim_{x \to 0} \frac{\tan 3x}{\sin 2x}$.

[2 marks]

Solution

Multiply and divide to create the standard patterns:

$$= \lim_{x \to 0} \frac{\frac{\tan 3x}{3x} \cdot 3x}{\frac{\sin 2x}{2x} \cdot 2x} = \frac{1 \cdot 3}{1 \cdot 2} = \frac{3}{2}$$

∴ 3/2

5

Evaluate $\lim_{x \to \infty} \frac{5x-3}{2x+7}$.

[2 marks]

Solution

Divide all terms by $x$ (highest power):

$$= \lim_{x \to \infty} \frac{5 - \frac{3}{x}}{2 + \frac{7}{x}} = \frac{5-0}{2+0} = \frac{5}{2}$$

∴ 5/2

6

Evaluate $\lim_{x \to 0} \frac{e^{5x}-e^{2x}}{x}$.

[2 marks]

Solution

Split the numerator:

$$= \lim_{x \to 0} \frac{(e^{5x}-1)-(e^{2x}-1)}{x}$$

$$= 5 \cdot \lim_{x \to 0}\frac{e^{5x}-1}{5x} - 2 \cdot \lim_{x \to 0}\frac{e^{2x}-1}{2x} = 5(1) - 2(1) = 3$$

∴ 3

7

Evaluate $\lim_{x \to 0} \frac{\ln(1+4x)}{x}$.

[2 marks]

Solution

Multiply and divide by 4 to match the standard pattern:

$$= 4 \cdot \lim_{x \to 0} \frac{\ln(1+4x)}{4x} = 4 \times 1 = 4$$

∴ 4

8

Evaluate $\lim_{x \to 0} \frac{1-\cos 4x}{x^2}$.

[2 marks]

Solution

Use $1-\cos 4x = 2\sin^2 2x$:

$$= \lim_{x \to 0} \frac{2\sin^2 2x}{x^2} = 2 \cdot \left(\frac{\sin 2x}{x}\right)^2 = 2 \cdot \left(2 \cdot \frac{\sin 2x}{2x}\right)^2 = 2 \cdot 4 \cdot 1^2 = 8$$

∴ 8

9

Evaluate $\lim_{x \to \pi/2} \sin x$.

[2 marks]

Solution

Sine is a continuous function everywhere. Direct substitution:

$$\lim_{x \to \pi/2} \sin x = \sin\left(\frac{\pi}{2}\right) = 1$$

∴ 1

10

Evaluate $\lim_{x \to 0} \frac{a^x - b^x}{x}$.

[2 marks]

Solution

Split the numerator:

$$= \lim_{x \to 0} \frac{(a^x-1)-(b^x-1)}{x} = \lim_{x \to 0}\frac{a^x-1}{x} - \lim_{x \to 0}\frac{b^x-1}{x}$$

$$= \ln a - \ln b = \ln\left(\frac{a}{b}\right)$$

∴ $\ln(a/b)$

Long Questions (4 Marks Each)

1

State and prove $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ using the Squeeze Theorem.

[4 marks]

Solution

Statement: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ where $\theta$ is in radians.

Proof: Consider a unit circle ($r=1$). For small $\theta > 0$, compare three areas:
• Area of $\triangle OPL = \frac{1}{2}\sin\theta\cos\theta$
• Area of sector $OPA = \frac{1}{2}\theta$
• Area of $\triangle OQA = \frac{1}{2}\tan\theta$

These satisfy: $\frac{1}{2}\sin\theta\cos\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$

Dividing by $\frac{1}{2}\sin\theta > 0$:
$$\cos\theta < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$$

Taking reciprocals (and reversing inequalities):
$$\cos\theta < \frac{\sin\theta}{\theta} < \frac{1}{\cos\theta}$$

As $\theta \to 0^+$: $\cos\theta \to 1$ and $\frac{1}{\cos\theta} \to 1$.

By the Squeeze Theorem: $\lim_{\theta \to 0^+} \frac{\sin\theta}{\theta} = 1$.

The result also holds for $\theta \to 0^-$ since $\frac{\sin(-\theta)}{-\theta} = \frac{\sin\theta}{\theta}$.

Therefore $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. $\blacksquare$

∴ Proved using Squeeze Theorem with triangle-sector area inequalities

2

Prove $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$ for positive integer $n$.

[4 marks]

Solution

Proof: We use the algebraic identity:
$$x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1})$$
This second factor has exactly $n$ terms.

Substitute into the limit:
$$\lim_{x \to a} \frac{(x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})}{x-a}$$

Since $x \to a$ means $x \neq a$, the factor $(x-a) \neq 0$. Cancel it:
$$= \lim_{x \to a}\; (x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1})$$

Now substitute $x = a$. Each of the $n$ terms becomes $a^{n-1}$:
$$= \underbrace{a^{n-1}+a^{n-1}+\cdots+a^{n-1}}_{n \text{ terms}} = na^{n-1} \quad \blacksquare$$

∴ Proved by algebraic factorisation of xⁿ − aⁿ

3

Evaluate $\lim_{x \to 2} \frac{x-\sqrt{8-x^2}}{\sqrt{x^2+12}-4}$.

[4 marks]

Solution

Direct substitution gives 0/0. Use double rationalisation.

Step 1 — Rationalise numerator: multiply by $(x+\sqrt{8-x^2})$:
$x - \sqrt{8-x^2}$ → $x^2 - (8-x^2) = 2x^2-8 = 2(x^2-4)$

Step 2 — Rationalise denominator: multiply by $(\sqrt{x^2+12}+4)$:
$\sqrt{x^2+12}-4$ → $x^2+12-16 = x^2-4$

Step 3 — Combined:
$$= \lim_{x \to 2} \frac{2(x^2-4)(\sqrt{x^2+12}+4)}{(x^2-4)(x+\sqrt{8-x^2})}$$

Cancel $(x^2-4)$:
$$= \lim_{x \to 2} \frac{2(\sqrt{x^2+12}+4)}{x+\sqrt{8-x^2}} = \frac{2(\sqrt{16}+4)}{2+\sqrt{4}} = \frac{2(8)}{4} = 4$$

∴ 4

4

Evaluate $\lim_{x \to \pi/2} \frac{1-\sin x}{(\pi/2-x)^2}$.

[4 marks]

Solution

Substitution: Let $y = \frac{\pi}{2} - x$. As $x \to \frac{\pi}{2}$, $y \to 0$.

Also: $\sin x = \sin\left(\frac{\pi}{2}-y\right) = \cos y$.

The limit becomes:
$$\lim_{y \to 0} \frac{1-\cos y}{y^2}$$

Apply $1-\cos y = 2\sin^2(y/2)$:
$$= \lim_{y \to 0} \frac{2\sin^2(y/2)}{y^2} = \frac{2}{4} \cdot \lim_{y \to 0}\left(\frac{\sin(y/2)}{y/2}\right)^2 = \frac{1}{2} \cdot 1^2 = \frac{1}{2}$$

∴ 1/2

5

Evaluate $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$.

[4 marks]

Solution

Express in terms of sin and cos:
$$\frac{\tan x - \sin x}{x^3} = \frac{\frac{\sin x}{\cos x}-\sin x}{x^3} = \frac{\sin x\left(\frac{1}{\cos x}-1\right)}{x^3} = \frac{\sin x(1-\cos x)}{x^3 \cos x}$$

Split into known limits:
$$= \frac{\sin x}{x} \cdot \frac{1-\cos x}{x^2} \cdot \frac{1}{\cos x}$$

Apply limits:
$$= 1 \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2}$$

∴ 1/2

6

If $\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to k} \frac{x^3-k^3}{x^2-k^2}$, find $k$.

[4 marks]

Solution

LHS: $\lim_{x \to 1}\frac{x^2-1}{x-1} = \lim_{x\to 1}\frac{(x-1)(x+1)}{x-1} = \lim_{x\to 1}(x+1) = 2$

RHS: $\lim_{x \to k}\frac{x^3-k^3}{x^2-k^2}$

Factor: $\frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}$. Cancel $(x-k)$:
$$= \lim_{x \to k}\frac{x^2+xk+k^2}{x+k} = \frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}$$

Set LHS = RHS:
$2 = \frac{3k}{2} \Rightarrow k = \frac{4}{3}$

∴ k = 4/3

7

Evaluate $\lim_{x \to a} \frac{x^{1/3}-a^{1/3}}{x^{1/2}-a^{1/2}}$.

[4 marks]

Solution

Divide both numerator and denominator by $(x-a)$:

$$= \frac{\lim_{x \to a}\frac{x^{1/3}-a^{1/3}}{x-a}}{\lim_{x \to a}\frac{x^{1/2}-a^{1/2}}{x-a}}$$

Apply algebraic theorem to each (with $n=1/3$ and $n=1/2$):

$$= \frac{\frac{1}{3}a^{1/3-1}}{\frac{1}{2}a^{1/2-1}} = \frac{\frac{1}{3}a^{-2/3}}{\frac{1}{2}a^{-1/2}}$$

$$= \frac{2}{3} \cdot a^{-2/3+1/2} = \frac{2}{3} \cdot a^{-1/6} = \frac{2}{3a^{1/6}}$$

∴ $\frac{2}{3}a^{-1/6}$

8

Evaluate $\lim_{x \to 0} \frac{e^{\sin x}-1}{x}$.

[4 marks]

Solution

As $x \to 0$, also $\sin x \to 0$. Rewrite by multiplying and dividing by $\sin x$:

$$= \lim_{x \to 0} \frac{e^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x}$$

Let $t = \sin x$. As $x \to 0$, $t \to 0$. Apply standard formulas:

$$= \lim_{t \to 0}\frac{e^t-1}{t} \cdot \lim_{x \to 0}\frac{\sin x}{x} = 1 \cdot 1 = 1$$

∴ 1

9

Evaluate $\lim_{x \to 0} \frac{\cos 3x - \cos 5x}{x^2}$.

[4 marks]

Solution

Use the sum-to-product identity: $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$

With $A=3x$ and $B=5x$:
$$\cos 3x - \cos 5x = -2\sin\frac{8x}{2}\sin\frac{-2x}{2} = -2\sin 4x \sin(-x) = 2\sin 4x \sin x$$

$$\lim_{x \to 0}\frac{2\sin 4x \sin x}{x^2} = 2 \cdot \frac{\sin 4x}{x} \cdot \frac{\sin x}{x}$$

$$= 2 \cdot 4 \cdot \frac{\sin 4x}{4x} \cdot \frac{\sin x}{x} = 2 \cdot 4 \cdot 1 \cdot 1 = 8$$

∴ 8

10

Check the continuity of $f(x) = \begin{cases} x+2 & x \le 1 \\ 3x & x > 1 \end{cases}$ at $x = 1$.

[4 marks]

Solution

Check all three conditions at $x = 1$:

Condition 1 — f(1) exists:
Since $x = 1$ falls in the first piece ($x \le 1$): $f(1) = 1+2 = 3$. ✓

Condition 2 — Limit exists:
LHL: $\lim_{x \to 1^-}(x+2) = 1+2 = 3$
RHL: $\lim_{x \to 1^+}(3x) = 3(1) = 3$
LHL = RHL = 3. ✓ Limit exists = 3.

Condition 3 — Limit = f(1):
$\lim_{x \to 1} f(x) = 3 = f(1) = 3$. ✓

All three conditions hold. $f$ is continuous at $x = 1$.

∴ f is continuous at x = 1 (all 3 conditions satisfied)

Past NEB Board Questions

1

State and prove $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.

NEB 2079 [4 marks]

Solution

See the full proof in Long Answer Q1 above. The key steps are:

1. Draw unit circle; establish area inequality: Area △OPL < Area sector OPA < Area △OQA.
2. Write: $\frac{1}{2}\sin\theta\cos\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$.
3. Divide by $\frac{1}{2}\sin\theta$: $\cos\theta < \frac{\theta}{\sin\theta} < \frac{1}{\cos\theta}$.
4. Invert: $\cos\theta < \frac{\sin\theta}{\theta} < \frac{1}{\cos\theta}$.
5. As $\theta\to 0$: both bounds → 1. By Squeeze Theorem: limit = 1. $\square$

∴ Proved by Squeeze Theorem — see full derivation above

2

Evaluate $\lim_{x \to 0} \frac{\cos 3x - \cos 5x}{x^2}$.

NEB 2078 [4 marks]

Solution

Apply sum-to-product: $\cos 3x - \cos 5x = 2\sin 4x \sin x$.

$$\lim_{x \to 0}\frac{2\sin 4x \sin x}{x^2} = 2 \cdot \lim_{x\to 0}\frac{\sin 4x}{4x}\cdot 4 \cdot \lim_{x\to 0}\frac{\sin x}{x} = 2 \cdot 1 \cdot 4 \cdot 1 = 8$$

∴ 8

3

Prove $\lim_{x \to a}\frac{x^n-a^n}{x-a} = na^{n-1}$ for positive integer $n$.

NEB 2077 [4 marks]

Solution

Use identity $x^n - a^n = (x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$.

Cancel $(x-a)$ (valid since $x\neq a$ in a limit).

Substitute $x=a$: each of the n terms = $a^{n-1}$. Sum = $na^{n-1}$. $\square$

∴ Proved — see full derivation in Long Answer Q2 above

4

Evaluate $\lim_{x \to 0} \frac{a^x - b^x}{x}$.

NEB 2076 [3 marks]

Solution

$$= \lim_{x \to 0}\frac{(a^x-1)-(b^x-1)}{x} = \lim_{x\to 0}\frac{a^x-1}{x} - \lim_{x\to 0}\frac{b^x-1}{x} = \ln a - \ln b = \ln\frac{a}{b}$$

∴ $\ln(a/b)$

Chapter Test — Full Mixed Paper

Attempt in 40 minutes without notes. Mirrors the NEB exam format.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

$\lim_{x \to 0} \frac{\sin 5x}{x}$ is:

Correct!

Incorrect. Correct: 5

Solution

$5 \cdot \lim_{x\to 0}\frac{\sin 5x}{5x} = 5 \times 1 = 5$

∴ 5

2

$\lim_{x \to \infty} \frac{2x+1}{x}$ is:

Correct!

Incorrect. Correct: 2

Solution

$\lim_{x\to\infty}\frac{2x+1}{x} = \lim_{x\to\infty}\left(2+\frac{1}{x}\right) = 2+0 = 2$

∴ 2

3

$\lim_{x \to 0} \frac{e^{2x}-1}{x}$ is:

Correct!

Incorrect. Correct: 2

Solution

$2 \cdot \lim_{x\to 0}\frac{e^{2x}-1}{2x} = 2 \times 1 = 2$

∴ 2

2 Section B — Short Questions (2 Marks Each)

1

Evaluate $\lim_{x \to 3} \frac{x^2-9}{x-3}$.

[2 marks]

Solution

$\frac{(x-3)(x+3)}{x-3} = x+3 \to 3+3 = 6$

∴ 6

2

Evaluate $\lim_{x \to 0} \frac{\ln(1+7x)}{x}$.

[2 marks]

Solution

$7 \cdot \lim_{x\to 0}\frac{\ln(1+7x)}{7x} = 7 \times 1 = 7$

∴ 7

3

Evaluate $\lim_{x \to 0} \frac{\tan x}{x}$.

[2 marks]

Solution

$\lim_{x\to 0}\frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \times 1 = 1$

∴ 1

3 Section C — Long Questions (4 Marks Each)

1

Prove $\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1$.

[4 marks]

Solution

Use unit circle area inequality. Area of △OPL < sector OPA < △OQA gives $\cos\theta < \frac{\sin\theta}{\theta} < \frac{1}{\cos\theta}$. As $\theta\to 0$, both bounds → 1. By Squeeze Theorem, limit = 1. $\square$

∴ Proved by Squeeze Theorem

2

Evaluate $\lim_{x \to \infty}(\sqrt{x+1}-\sqrt{x})$.

[4 marks]

Solution

Rationalise: $\frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \frac{1}{\sqrt{x+1}+\sqrt{x}}$. As $x\to\infty$, denominator $\to\infty$, so limit $= 0$.

∴ 0

One-Page Revision Cheat Sheet

1

7 Indeterminate Forms: 0/0, ∞/∞, ∞−∞, 0·∞, 0⁰, 1^∞, ∞⁰. Note: 0^∞ = 0, NOT indeterminate.

2

Algebraic: $(x^n-a^n)/(x-a) \to na^{n-1}$. Works for all rational n.

3

Trig: $\sin\theta/\theta \to 1$ and $\tan\theta/\theta \to 1$ as $\theta\to 0$ (radians only!).

4

1−cos trick: $1-\cos x = 2\sin^2(x/2)$, so $(1-\cos x)/x^2 \to 1/2$.

5

Exponential: $(e^x-1)/x \to 1$. $(a^x-1)/x \to \ln a$.

6

Logarithmic: $\ln(1+x)/x \to 1$. For kx: multiply-divide by k.

7

At infinity: Divide by highest power. Same degree → ratio of coefficients. Lower degree → 0. Higher degree → ±∞.

8

Mirror trick: $\sin(kx)/x = k \cdot \sin(kx)/(kx) \to k$.

9

Degrees trap: $\sin x°/x \to \pi/180$, NOT 1. Always convert to radians.

10

Continuity — 3 conditions: f(a) exists AND lim f(x) exists AND lim = f(a).

11

LHL = RHL is necessary for limit existence. Check both for piecewise/|x| functions.

12

Types of discontinuity: Removable (hole), Jump (LHL≠RHL), Infinite (limit=±∞), Oscillatory.

13

Standard chain: Try direct substitution → if 0/0: factor/rationalise → if trig: mirror trick → if ∞: divide by highest power.