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

Chapter Overview

This chapter introduces the core concept of Limits and Continuity, which serves as the foundational gateway to Calculus (Derivatives and Integrals). In the NEB Class 11 syllabus, this topic is crucial as it transitions from static algebra to dynamic mathematical analysis, testing your ability to evaluate function behavior near singular points.

What this chapter covers

Neighborhoods and Deleted NeighborhoodsLeft Hand Limits (LHL) & Right Hand Limits (RHL)Existence of a LimitThree-Step Continuity CriteriaRemovable & Infinite DiscontinuitiesEvaluating Absolute Value LimitsFinding Unknown Constants for ContinuityJump of a Function

Learning Objectives

1

Distinguish between neighborhoods and deleted neighborhoods geometrically and algebraically

2

Evaluate Left Hand Limits (LHL) and Right Hand Limits (RHL) for piecewise and absolute value functions

3

Determine limit existence at a specific point by verifying if LHL equals RHL

4

Apply the three-step algebraic test to verify the continuity of a function at a boundary

5

Identify, classify, and mathematically resolve removable and jump discontinuities

6

Formulate and solve algebraic equations to find unknown parameters (a, b, k) that ensure continuity

The One Rule That Solves Continuity

Key Insight

To establish continuity at any point $x = a$, you must perform the L-R-F check: the Left-Hand Limit must equal the Right-Hand Limit, and both must equal the finite Functional Value.

$$\text{LHL} = \text{RHL} = f(a)$$

If paths from both sides meet, the limit exists. If they meet exactly at the solid plotted point $f(a)$, the function is continuous. If they do not, a gap or hole remains.

Key Definitions

Neighborhood

For a point $a$, the open interval $(a-\delta, a+\delta)$ is called its $\delta$-neighborhood. It contains all real numbers whose distance from $a$ is strictly less than $\delta$.

Deleted Neighborhood

The $\delta$-neighborhood of $a$ excluding the point $a$ itself, denoted by $(a-\delta, a+\delta) - \{a\}$ or $0 < |x-a| < \delta$. This is the mathematical domain where limits are evaluated, as they ignore the exact point $a$.

Limit of a Function

The value $l$ that a function $f(x)$ approaches as the independent variable $x$ gets arbitrarily close to $a$. Evaluated using the formal $\epsilon-\delta$ framework.

$$0 < |x-a| < \delta \Rightarrow |f(x)-l| < \epsilon$$

Continuous Function

A function is continuous at a point $x=a$ if its graph can be drawn through that point without lifting the pencil. Algebraically, the limit value as $x \to a$ must match the defined functional value $f(a)$.

Left Hand Limit (LHL)

The limiting value of $f(x)$ as $x$ approaches $a$ from the left side (from values strictly smaller than $a$, where $x < a$).

Right Hand Limit (RHL)

The limiting value of $f(x)$ as $x$ approaches $a$ from the right side (from values strictly larger than $a$, where $x > a$).

Limit Existence vs Continuity

The fundamental difference lies in whether we check the behavior exactly at the point, or merely close to the point.

Limit Existence

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

Near, not at — Cares only about the behavior in the deleted neighborhood of $a$.

Does not require $f(a)$ to be defined or even exist.

Keywords: evaluate, limit exists, find limit

Example: $f(x) = \frac{x^2-16}{x-4}$ has a limit of $8$ at $x=4$, even though $f(4)$ is indeterminate ($0/0$).

Is a necessary pre-requisite for continuity.

VS

Continuity

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

Both near and at — Merges the limit behavior with the actual functional value at $x=a$.

Strictly requires $f(a)$ to be defined as a finite real value.

Keywords: test continuity, continuous everywhere, find constants

Example: $f(x) = x^2$ is continuous at $x=4$ because the limit is $16$ and $f(4) = 16$.

Graphically, there is no hole, jump, or vertical asymptote at $x=a$.

Memory Trick: Limit is about the destination. Continuity is about making sure the road doesn't have a bridge collapse at that destination.

Formula Sheet — Formal Calculus Rules

Formal Limit Definition

0 < |x-a| < \delta \Rightarrow |f(x)-l| < \epsilon

Epsilon-delta criteria for the existence of limit $l$ as $x \to a$.

1Let $\epsilon > 0$ represent any small vertical boundary around the limit value $l$.

2To prove the limit exists, we must find a horizontal boundary $\delta > 0$ around $a$ such that every $x$ value in the deleted interval maps inside the vertical boundary.

3The term $0 < |x-a|$ explicitly excludes $x = a$ itself, ensuring we only evaluate behaviors in the neighborhood.

Left Hand Limit

f(a^-) = \lim_{h \to 0} f(a-h)

Used to evaluate the limit from values smaller than $a$ ($x < a$).

1We substitute $x = a - h$, where $h$ is a positive real number ($h > 0$).

2As $x$ approaches $a$ from the left side, the gap $h$ must shrink down to $0$.

3Evaluating $\lim_{h \to 0} f(a-h)$ yields the Left Hand Limit.

Right Hand Limit

f(a^+) = \lim_{h \to 0} f(a+h)

Used to evaluate the limit from values larger than $a$ ($x > a$).

1We substitute $x = a + h$, where $h$ is a positive real number ($h > 0$).

2As $x$ approaches $a$ from the right side, the gap $h$ must shrink down to $0$.

3Evaluating $\lim_{h \to 0} f(a+h)$ yields the Right Hand Limit.

Jump of a Function

\text{Jump} = |f(a^+) - f(a^-)|

Measures the size of the gap at a jump discontinuity.

1In piecewise boundaries, LHL and RHL may both approach finite, but different, values.

2This creates a non-overlapping vertical break (a jump) on the graph.

3The distance of this gap is calculated by taking the absolute value of the difference between the Right Hand Limit $f(a^+)$ and the Left Hand Limit $f(a^-)$.

Continuity from Definition

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

The ultimate condition for a function to be continuous at $x = a$.

1From the formal definition of continuity: for any $\epsilon > 0$, there exists $\delta > 0$ such that $|x-a| < \delta \Rightarrow |f(x)-f(a)| < \epsilon$.

2Notice that $|x-a| < \delta$ does not have a lower bound of $0$, meaning $x = a$ is included in the domain.

3As $x$ gets close to $a$, the distance between the values of $f(x)$ and $f(a)$ must approach $0$, which algebraically states that the limit matches the functional value: $\lim_{x \to a} f(x) = f(a)$.

Continuous or Discontinuous?

Follow this decision-making tree to systematically analyze limits and continuity at any given point.

Is the functional value f(a) defined?

Yes

Continue to check the limit ↓

No

Discontinuous! (Either Removable or Infinite Discontinuity)

Do the Left Hand Limit and Right Hand Limit match? (LHL = RHL)

Yes

Limit exists! Continue to final verification ↓

No

Discontinuous! (Jump Discontinuity or Non-Existent Limit)

Does the Limit equal the Functional Value? (Limit = f(a))

Yes

Continuous! The function has no breaks or holes at x=a.

No

Discontinuous! (Removable Discontinuity — Redefining f(a) can fix this)

The Absolute Value Piecewise Trick

Trick: Limits involving $|x-a|$ must be split into two separate algebraic cases using the definition of absolute value.

Steps to Remember:

Step 1: For the Left Hand Limit ($x \to a^-$), we know $x < a$. Therefore, replace the term $|x-a|$ with $-(x-a)$.
Step 2: For the Right Hand Limit ($x \to a^+$), we know $x > a$. Therefore, replace the term $|x-a|$ with $+(x-a)$.
Step 3: Simplify both algebraic expressions by canceling the common terms, then evaluate the resulting constant limits.
Step 4: Compare LHL and RHL. If they do not match, state that the limit does not exist.

Mnemonic: LHL gets the negative block: -(x-a). RHL gets the positive block: +(x-a).

Common Mistakes

These conceptual mistakes are frequently encountered in NEB mathematics exams. Ensure you avoid them.

Confusing limit existence with continuity — A limit can exist (LHL = RHL) while the function remains discontinuous because $f(a)$ is undefined or equals a different value. Always check all three L-R-F conditions.
Evaluating 1/0 as a finite limit — Getting $1/0$ means the function shoots off to infinity. State that the limit does not exist as a finite real number; do not write it as a valid numerical limit.
Forgetting absolute value sign changes — Students often evaluate $\lim_{x \to a} \frac{|x-a|}{x-a}$ as $1$ without realizing that approaching from the left yields $-1$.
Plugging in boundaries blindly — In piecewise questions, always evaluate limits from both sides of a boundary value instead of just substituting the boundary into one of the expressions.

NEB Exam Traps — High-Yield Edge Cases

NEB Exam Warning

Board examiners design questions targeting specific algebraic pitfalls. Keep these checkpoints in mind.

Piecewise discontinuity transitions — Check the exact domain inequalities ($<$ vs $\le$). If a value is excluded from both intervals, the function is automatically discontinuous because $f(a)$ is undefined.
Indeterminate forms of trigonometric limits — When solving $\lim_{x \to 0} \frac{\sin 5x}{x}$, do not simply substitute $0/0$. Use the identity $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ by multiplying and dividing by the matching coefficient.
Solving for two variables (a and b) — If a function has two transition boundaries, you must set up and solve two distinct systems of equations using the continuity condition at both transition points.
Jump of a function — Jump calculations require the absolute difference: $|\text{RHL} - \text{LHL}|$. It must always be a positive quantity representing physical distance.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Question uses curly braces to define different formulas on different intervals	Piecewise analysis: Calculate LHL, RHL, and f(a) separately using the boundaries.
Limit has an absolute value term like \|x - a\| in the numerator or denominator	Absolute value trick: Split into x < a (negative) and x > a (positive) limits.
Question asks you to 'find k' or 'find a and b' given that the function is continuous	Set LHL = RHL = f(a) at the boundary points, then solve the algebraic equations.
Limit evaluation yields a 0/0 form	Factorize the polynomials, rationalize the surds, or apply trigonometric limit identities.
Limit evaluation yields a non-zero number divided by 0	State that the limit does not exist as a finite real value (infinite limit).
Question asks for the 'jump' of a function at a boundary	Calculate the LHL and RHL, then evaluate the absolute difference: \|RHL - LHL\|.

Continuous vs. Discontinuous Functions

The left plot shows a continuous function where the limit as $x \to a$ matches the solid point $f(a)$. The right plot shows a function with a removable discontinuity (a hole) at $x = a$, while the actual value $f(a)$ is plotted elsewhere. This visualizes why matching the limit is required for continuity.

A continuous curve has no breaks. A curve with a removable discontinuity has a hole at the limit point, with the point defined elsewhere.

Solved Examples (Easy → Hard)

1

Find the limit of $f(x)$ at $x=1$ if $f(x) = \begin{cases} 4x+1 & \text{for } x < 1 \\ 5 & \text{for } x \ge 1 \end{cases}$

Easy

Solution

To find the limit, we evaluate the Left-Hand and Right-Hand limits at $x = 1$.

1. Left-Hand Limit (LHL):
For $x < 1$, we use $f(x) = 4x+1$:
$$\text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1} (4x+1) = 4(1)+1 = 5$$

2. Right-Hand Limit (RHL):
For $x \ge 1$, we use $f(x) = 5$:
$$\text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1} (5) = 5$$

3. Conclusion:
Since $\text{LHL} = \text{RHL} = 5$, the limit exists and is equal to $5$.

∴ 5

2

Test the continuity of the function $f(x) = \frac{x^2 - 16}{x-4}$ at $x=4$.

Medium

Solution

We apply the three-step criteria of continuity at $x = 4$.

Step 1 — Functional Value:
$$f(4) = \frac{4^2 - 16}{4-4} = \frac{0}{0}$$
Since $\frac{0}{0}$ is an indeterminate form, $f(4)$ is undefined.

Step 2 — Limit Value:
Evaluate the limit by factorizing the numerator:
$$\lim_{x \to 4} \frac{x^2 - 16}{x-4} = \lim_{x \to 4} \frac{(x-4)(x+4)}{x-4}$$
Cancel the common factor $(x-4)$:
$$\lim_{x \to 4} (x+4) = 4 + 4 = 8$$

Step 3 — Verification:
Since the functional value $f(4)$ does not exist, the continuity condition $\lim_{x \to 4} f(x) = f(4)$ fails. Therefore, the function is discontinuous at $x = 4$.

∴ Discontinuous at x = 4

3

Evaluate $\lim_{x \to 2} \frac{|x-2|}{x-2}$ if it exists.

Hard

Solution

Since this limit involves an absolute value centered at the transition point $x=2$, we must evaluate the LHL and RHL independently.

1. Left-Hand Limit (LHL):
For $x < 2$, we have $|x-2| = -(x-2)$:
$$\text{LHL} = \lim_{x \to 2^-} \frac{-(x-2)}{x-2} = \lim_{x \to 2^-} (-1) = -1$$

2. Right-Hand Limit (RHL):
For $x > 2$, we have $|x-2| = (x-2)$:
$$\text{RHL} = \lim_{x \to 2^+} \frac{(x-2)}{x-2} = \lim_{x \to 2^+} (1) = 1$$

3. Conclusion:
Since the Left-Hand Limit ($-1$) does not equal the Right-Hand Limit ($1$), the limit does not exist.

∴ Does not exist

MCQ Practice

Mixed Level — MCQ

1

A function f(x) is continuous at x = a if:

Correct!

Incorrect. Correct: All of the above

Solution

Continuity at a point requires three criteria: 1) $f(a)$ is defined, 2) $\lim_{x \to a} f(x)$ exists, and 3) the limit value equals the functional value. Therefore, all conditions must hold.

∴ All of the above

2

If f(x) = 1 / (x - 3), what is the value of the limit as x approaches 3?

Correct!

Incorrect. Correct: Does not exist

Solution

Evaluate both limits:
$$\lim_{x \to 3^-} \frac{1}{x-3} = -\infty$$
$$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$$
Since the values do not approach a single real number, the limit does not exist.

∴ Does not exist

3

The set (a - \delta, a + \delta) - {a} is mathematically classified as the:

Correct!

Incorrect. Correct: Deleted Neighborhood

Solution

By definition, a deleted neighborhood of a point $a$ is any neighborhood of $a$ with the point $a$ itself removed.

∴ Deleted Neighborhood

4

Determine the limit as x approaches 0 of the function f(x) = x^2 * cos(1/x).

Correct!

Incorrect. Correct: 0

Solution

Since $-1 \le \cos(1/x) \le 1$ for all $x \neq 0$, we can set up bounds:
$$-x^2 \le x^2 \cos(1/x) \le x^2$$
Applying the Squeeze Theorem as $x \to 0$:
$$\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} (x^2) = 0$$
Therefore, the limit is $0$.

∴ 0

5

A removable discontinuity at x = a can be resolved by:

Correct!

Incorrect. Correct: Redefining f(a) to equal the limit

Solution

A removable discontinuity occurs when $\lim_{x \to a} f(x) = l$ exists but $f(a) \neq l$. By resetting $f(a) = l$, the function becomes continuous.

∴ Redefining f(a) to equal the limit

6

The function f(x) = |x| / x at x = 0 has what type of discontinuity?

Correct!

Incorrect. Correct: Jump discontinuity

Solution

Calculate LHL and RHL at $x=0$:
$$\text{LHL} = \lim_{x \to 0^-} \frac{-x}{x} = -1$$
$$\text{RHL} = \lim_{x \to 0^+} \frac{x}{x} = 1$$
Since LHL and RHL are unequal finite constants, it is a jump discontinuity.

∴ Jump discontinuity

7

On which of the following domains are polynomial functions continuous?

Correct!

Incorrect. Correct: All real numbers \mathbb{R}

Solution

Every polynomial function $P(x) = a_n x^n + \dots + a_0$ is defined for all real numbers and satisfies $\lim_{x \to c} P(x) = P(c)$ for any real $c$. Hence, it is continuous on $\mathbb{R}$.

∴ All real numbers \mathbb{R}

8

If LHL = 5 and RHL = 5, but f(a) = 3, then f(x) is:

Correct!

Incorrect. Correct: Discontinuous

Solution

For continuity, we require $\lim_{x \to a} f(x) = f(a)$. Here, the limit is $5$, but $f(a) = 3$. Since $5 \neq 3$, the function is discontinuous.

∴ Discontinuous

9

Calculate the jump of a function at a boundary where LHL = -2 and RHL = 3.

Correct!

Incorrect. Correct: 5

Solution

Use the jump formula:
$$\text{Jump} = |\text{RHL} - \text{LHL}| = |3 - (-2)| = |5| = 5$$

∴ 5

10

Evaluate the limit as x approaches 0 of sin(1/x).

Correct!

Incorrect. Correct: Non-existent due to oscillation

Solution

Because the function oscillates infinitely between $-1$ and $1$ as $x$ approaches $0$, it does not approach any single unique value. Thus, the limit does not exist.

∴ Non-existent due to oscillation

Short Questions (2 Marks Each)

1

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

[2 marks]

Solution

Direct substitution yields the indeterminate form $\frac{0}{0}$. We factorize the numerator:

$$\lim_{x \to 3} \frac{(x-3)(x+3)}{x-3}$$
Cancel the common factor $(x-3)$ for $x \neq 3$:
$$\lim_{x \to 3} (x+3) = 3 + 3 = 6$$

∴ 6

2

Define Left Hand Limit in $\epsilon-\delta$ notation.

[2 marks]

Solution

The Left Hand Limit $\lim_{x \to a^-} f(x) = l$ if for every $\epsilon > 0$, there exists a corresponding $\delta > 0$ such that:

$$a-\delta < x < a \Rightarrow |f(x) - l| < \epsilon$$
This restricts the inputs $x$ to values strictly smaller than $a$.

∴ a - \delta < x < a \Rightarrow |f(x) - l| < \epsilon

3

Test if $f(x) = 5x-2$ is continuous at $x=2$.

[2 marks]

Solution

Verify the continuity conditions at $x = 2$:

1. Functional Value: $f(2) = 5(2) - 2 = 8$
2. Limit Value: $\lim_{x \to 2} (5x-2) = 5(2)-2 = 8$
3. Verification: Since $\lim_{x \to 2} f(x) = f(2) = 8$, the function is continuous.

∴ Continuous

4

Find the LHL and RHL for $f(x) = \begin{cases} 2 & x \le 0 \\ 3 & x > 0 \end{cases}$ at $x=0$.

[2 marks]

Solution

Evaluate both limits at the transition point $x = 0$:

Left Hand Limit:
$$\text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0} 2 = 2$$
Right Hand Limit:
$$\text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0} 3 = 3$$

∴ LHL = 2, RHL = 3

5

Is $f(x) = \tan x$ continuous at $x = \pi/2$? Why?

[2 marks]

Solution

At $x = \pi/2$, the functional value $f(\pi/2) = \tan(\pi/2)$ is undefined (approaches infinity).

Since the functional value does not exist as a finite real number, the function is discontinuous.

∴ Discontinuous

6

Find the jump of $f(x) = \begin{cases} x+1 & x < 1 \\ 4-x & x \ge 1 \end{cases}$ at $x=1$.

[2 marks]

Solution

First, find the Left-Hand Limit and Right-Hand Limit at $x = 1$:

$$\text{LHL} = \lim_{x \to 1^-} (x+1) = 1+1 = 2$$
$$\text{RHL} = \lim_{x \to 1^+} (4-x) = 4-1 = 3$$
Now calculate the absolute difference:
$$\text{Jump} = |\text{RHL} - \text{LHL}| = |3-2| = 1$$

∴ 1

7

What is a 'Deleted Neighborhood'? Give an example centered at $a=5$ with radius $\delta = 0.1$.

[2 marks]

Solution

A deleted neighborhood of a point $a$ is an open interval centered at $a$ that excludes $a$ itself.

For $a = 5$ and $\delta = 0.1$:
$$(5 - 0.1, 5 + 0.1) - \{5\} = (4.9, 5.1) - \{5\}$$
This can be expressed using inequalities as $0 < |x - 5| < 0.1$.

∴ (4.9, 5.1) - {5}

8

Evaluate $\lim_{x \to 0} x \sin(1/x)$.

[2 marks]

Solution

We bound the oscillating function $\sin(1/x)$ between $-1$ and $1$:
$$-1 \le \sin(1/x) \le 1$$
Multiply all parts by $x$ (for $x > 0$, similar signs apply for $x < 0$):
$$-x \le x \sin(1/x) \le x$$
Since $\lim_{x \to 0} (-x) = 0$ and $\lim_{x \to 0} x = 0$, by the Squeeze Theorem:
$$\lim_{x \to 0} x \sin(1/x) = 0$$

∴ 0

9

If $\lim_{x \to a} f(x)$ exists and is finite, but $f(a)$ is undefined, what type of discontinuity is present?

[2 marks]

Solution

Since the limit exists at $x = a$, the paths from the left and right meet at a single point. However, because $f(a)$ is undefined, there is a missing point (a hole). This is classified as a removable discontinuity.

∴ Removable Discontinuity

10

State the properties of sum and product for continuous functions.

[2 marks]

Solution

If two functions $f(x)$ and $g(x)$ are both continuous at a point $x = a$, then:

1. Their sum $(f + g)(x)$ is continuous at $x = a$.
2. Their product $(f \cdot g)(x)$ is continuous at $x = a$.

∴ f+g and f*g are continuous

Long Questions (3 Marks Each)

1

Find the value of $k$ such that the function $f(x) = \begin{cases} kx^2 + 1 & x \le 2 \\ 3x - 1 & x > 2 \end{cases}$ is continuous at $x=2$.

[3 marks]

Solution

For the function to be continuous at $x = 2$, we require $\text{LHL} = \text{RHL} = f(2)$.

Step 1 — Functional Value:
$$f(2) = k(2)^2 + 1 = 4k + 1$$

Step 2 — Left Hand Limit (LHL):
$$\text{LHL} = \lim_{x \to 2^-} (kx^2 + 1) = 4k + 1$$

Step 3 — Right Hand Limit (RHL):
$$\text{RHL} = \lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5$$

Step 4 — Equate and Solve:
Set LHL = RHL:
$$4k + 1 = 5$$
$$4k = 4 \Rightarrow k = 1$$

∴ k = 1

2

Determine the values of $a$ and $b$ so that the function $f(x)$ is continuous everywhere: $f(x) = \begin{cases} 2x+1 & x < 1 \\ a & x = 1 \\ bx^2 + 3 & x > 1 \end{cases}$

[3 marks]

Solution

For continuity at $x = 1$, we require $\text{LHL} = \text{RHL} = f(1)$.

Step 1 — Evaluate LHL:
$$\text{LHL} = \lim_{x \to 1^-} (2x+1) = 2(1)+1 = 3$$

Step 2 — Evaluate f(1):
$$f(1) = a$$
Set $\text{LHL} = f(1) \Rightarrow a = 3$.

Step 3 — Evaluate RHL:
$$\text{RHL} = \lim_{x \to 1^+} (bx^2 + 3) = b(1)^2 + 3 = b+3$$
Set $\text{LHL} = \text{RHL}$:
$$3 = b + 3 \Rightarrow b = 0$$

Answer: $a = 3$, $b = 0$.

∴ a = 3, b = 0

3

Examine the continuity of $f(x) = \frac{|x-5|}{x-5}$ at $x=5$. If discontinuous, calculate the jump.

[3 marks]

Solution

Evaluate the functional value and the limits at $x = 5$:

Step 1 — Functional Value:
$$f(5) = \frac{|5-5|}{5-5} = \frac{0}{0} \quad \text{(Undefined)}$$

Step 2 — Left Hand Limit (x < 5):
Replace $|x-5|$ with $-(x-5)$:
$$\text{LHL} = \lim_{x \to 5^-} \frac{-(x-5)}{x-5} = -1$$

Step 3 — Right Hand Limit (x > 5):
Replace $|x-5|$ with $+(x-5)$:
$$\text{RHL} = \lim_{x \to 5^+} \frac{x-5}{x-5} = 1$$

Step 4 — Conclusion:
Since $f(5)$ is undefined and $\text{LHL} \neq \text{RHL}$, the function is discontinuous. The jump is:
$$\text{Jump} = |\text{RHL} - \text{LHL}| = |1 - (-1)| = 2$$

∴ Discontinuous, Jump = 2

4

Show that the function $f(x) = \begin{cases} \frac{\sin 5x}{x} & x \neq 0 \\ 5 & x = 0 \end{cases}$ is continuous at $x=0$.

[3 marks]

Solution

To verify continuity at $x = 0$, we perform the L-R-F check.

Step 1 — Functional Value:
$$f(0) = 5$$

Step 2 — Limit Value:
Evaluate the limit using the trigonometric identity $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \left(\frac{\sin 5x}{5x}\right) = 5 \cdot 1 = 5$$

Step 3 — Compare:
Since $\lim_{x \to 0} f(x) = f(0) = 5$, the function is continuous.

∴ Continuous

5

Let $f(x) = \begin{cases} x^2 - 1 & x < 3 \\ 10 & x = 3 \\ 2x+2 & x > 3 \end{cases}$. Test if continuous at $x=3$, and if not, show how to make it continuous.

[3 marks]

Solution

Step 1 — Evaluate LHL (x < 3):
$$\text{LHL} = \lim_{x \to 3^-} (x^2 - 1) = 3^2 - 1 = 8$$

Step 2 — Evaluate RHL (x > 3):
$$\text{RHL} = \lim_{x \to 3^+} (2x + 2) = 2(3) + 2 = 8$$
The limit exists and equals $8$.

Step 3 — Compare with f(3):
We are given $f(3) = 10$. Since $\lim_{x \to 3} f(x) = 8 \neq 10$, the function is discontinuous at $x = 3$.

How to make it continuous:
Redefine the point value to equal the limit: $f(3) = 8$.

∴ Discontinuous; make continuous by setting f(3) = 8

6

Prove using the limit definition why $\lim_{x \to 0} \frac{1}{x}$ does not exist.

[3 marks]

Solution

We evaluate LHL and RHL separately as $x \to 0$:

1. Left-Hand Limit:
As $x$ approaches $0$ from negative values, the value of $\frac{1}{x}$ becomes arbitrarily large in the negative direction:
$$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$

2. Right-Hand Limit:
As $x$ approaches $0$ from positive values, the value of $\frac{1}{x}$ becomes arbitrarily large in the positive direction:
$$\lim_{x \to 0^+} \frac{1}{x} = +\infty$$

Since the values do not approach a unique, finite real number, and the directions are opposite, the limit does not exist.

∴ Does not exist due to infinite limits

7

Find the Left Jump, Right Jump, and Total Jump for $f(x) = \begin{cases} x & x < 0 \\ 1 & x = 0 \\ 2-x & x > 0 \end{cases}$ at $x=0$.

[3 marks]

Solution

Calculate limits and compare with the point value $f(0) = 1$:

$$\text{LHL} = \lim_{x \to 0^-} x = 0$$
$$\text{RHL} = \lim_{x \to 0^+} (2-x) = 2$$

1. Left Jump: Difference between functional value and LHL:
$$\text{Left Jump} = |f(0) - \text{LHL}| = |1 - 0| = 1$$
2. Right Jump: Difference between RHL and functional value:
$$\text{Right Jump} = |\text{RHL} - f(0)| = |2 - 1| = 1$$
3. Total Jump: Difference between RHL and LHL:
$$\text{Total Jump} = |\text{RHL} - \text{LHL}| = |2 - 0| = 2$$

∴ Left=1, Right=1, Total=2

8

Discuss the continuity of the function $f(x) = \begin{cases} 3+2x & -3/2 \le x < 0 \\ 3-2x & 0 \le x < 3/2 \end{cases}$ at $x=0$.

[3 marks]

Solution

Perform the three-step check at $x = 0$:

1. Functional Value: Use the second interval containing $\le x$:
$$f(0) = 3 - 2(0) = 3$$

2. Left Hand Limit (x < 0):
$$\text{LHL} = \lim_{x \to 0^-} (3+2x) = 3 + 2(0) = 3$$

3. Right Hand Limit (x > 0):
$$\text{RHL} = \lim_{x \to 0^+} (3-2x) = 3 - 2(0) = 3$$

Since $\text{LHL} = \text{RHL} = f(0) = 3$, the function is continuous at $x = 0$.

∴ Continuous

9

Show that the function $f(x) = x^2$ is continuous on the interval $[1, 3]$.

[3 marks]

Solution

A function is continuous on a closed interval $[a, b]$ if it is continuous at every point in the open interval $(a, b)$, continuous from the right at $a$, and continuous from the left at $b$.

1. Inside the interval: For any $c \in (1, 3)$, we have:
$$\lim_{x \to c} x^2 = c^2 = f(c)$$
Thus, it is continuous at all internal points.

2. Right-hand boundary: At $x=1$:
$$\lim_{x \to 1^+} x^2 = 1^2 = 1 = f(1)$$
3. Left-hand boundary: At $x=3$:
$$\lim_{x \to 3^-} x^2 = 3^2 = 9 = f(3)$$
Since all three criteria are satisfied, the function is continuous on the interval.

∴ Continuous on the interval

10

Explain using mathematical examples the difference between a Removable Discontinuity and an Infinite Discontinuity.

[3 marks]

Solution

1. Removable Discontinuity: Occurs when the limit exists but does not equal the functional value.
Example: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.
The limit is $\lim_{x \to 2} (x+2) = 4$, but $f(2)$ is undefined. The gap is a single missing point which can be filled.

2. Infinite Discontinuity: Occurs when one or both of the one-sided limits go to $\pm\infty$.
Example: $f(x) = \frac{1}{x-2}$ at $x = 2$.
As $x \to 2^+$, $f(x) \to \infty$, creating a vertical asymptote. This cannot be fixed by redefining a single point.

∴ Removable: limit exists; Infinite: limit goes to infinity

Past NEB Board Questions

1

Examine the continuity of the function $f(x) = \frac{|x-5|}{x-5}$ at $x=5$. If discontinuous, find the jump.

NEB 2079 [3 marks]

Solution

Step 1: Check Functional Value
$$f(5) = \frac{|5-5|}{5-5} = \frac{0}{0} \quad \text{(Undefined)}$$

Step 2: Find LHL (x < 5)
Substitute $|x-5| = -(x-5)$:
$$\text{LHL} = \lim_{x \to 5^-} \frac{-(x-5)}{x-5} = -1$$

Step 3: Find RHL (x > 5)
Substitute $|x-5| = (x-5)$:
$$\text{RHL} = \lim_{x \to 5^+} \frac{x-5}{x-5} = 1$$

Step 4: Compute Jump
Since the LHL and RHL do not match, the limit does not exist, confirming discontinuity. The jump is:
$$\text{Jump} = |1 - (-1)| = 2$$

∴ Discontinuous, Jump = 2

2

Find the value of $k$ such that the function $f(x) = \begin{cases} kx^2 + 1 & x \le 2 \\ 3x - 1 & x > 2 \end{cases}$ is continuous at $x=2$.

NEB 2078 [3 marks]

Solution

For continuity at the transition point $x=2$, we require the Left-Hand Limit to equal the Right-Hand Limit.

1. Left-Hand Limit:
$$\text{LHL} = \lim_{x \to 2^-} (kx^2 + 1) = k(2)^2 + 1 = 4k + 1$$

2. Right-Hand Limit:
$$\text{RHL} = \lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5$$

3. Solve for k:
Set LHL = RHL:
$$4k + 1 = 5$$
$$4k = 4 \Rightarrow k = 1$$

∴ k = 1

3

Show that the function $f(x) = \begin{cases} \frac{\sin 5x}{x} & x \neq 0 \\ 5 & x = 0 \end{cases}$ is continuous at $x=0$.

NEB 2077 [3 marks]

Solution

Check the L-R-F continuity criteria at $x = 0$:

1. Functional Value:
$$f(0) = 5$$

2. Limit Value:
Evaluate the limit by matching the trigonometric denominator:
$$\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x}$$
Using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$$= 5 \cdot 1 = 5$$

3. Conclusion:
Since $\lim_{x \to 0} f(x) = f(0) = 5$, the function is continuous.

∴ Continuous

Chapter Test — Full Mixed Paper

Complete this examination-style test paper in 45 minutes under quiet exam conditions.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

If the limit of f(x) exists as x approaches a, then the value of the limit must be a:

Correct!

Incorrect. Correct: Unique finite real number

Solution

By the standard definition of limits, the limiting value must be a unique, finite real number.

∴ Unique finite real number

2

The function f(x) = |x| at x = 0 is:

Correct!

Incorrect. Correct: Continuous

Solution

$$f(0) = |0| = 0$$
$$\text{LHL} = \lim_{x \to 0^-} (-x) = 0$$
$$\text{RHL} = \lim_{x \to 0^+} (x) = 0$$
Since LHL = RHL = f(0), the function is continuous.

∴ Continuous

3

Find the jump of f(x) at x = a if f(a^+) = 10 and f(a^-) = 4.

Correct!

Incorrect. Correct: 6

Solution

$$\text{Jump} = |\text{RHL} - \text{LHL}| = |10 - 4| = 6$$

∴ 6

2 Section B — Short Questions (2 Marks Each)

1

Evaluate $\lim_{x \to 0} x^2 \cos(1/x)$.

[2 marks]

Solution

The function $\cos(1/x)$ is bounded between $-1$ and $1$.
Multiply by the positive term $x^2$:
$$-x^2 \le x^2 \cos(1/x) \le x^2$$
Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$, by the Squeeze Theorem:
$$\lim_{x \to 0} x^2 \cos(1/x) = 0$$

∴ 0

2

Show that the function $f(x) = \begin{cases} 2x+3 & x < 1 \\ 5 & x \ge 1 \end{cases}$ is continuous at $x=1$.

[2 marks]

Solution

Verify the continuity conditions at $x = 1$:
$$\text{LHL} = \lim_{x \to 1^-} (2x+3) = 2(1)+3 = 5$$
$$\text{RHL} = \lim_{x \to 1^+} (5) = 5$$
$$\text{Functional value } f(1) = 5$$
Since $\text{LHL} = \text{RHL} = f(1) = 5$, the function is continuous.

∴ Continuous

3 Section C — Long Questions (3 Marks Each)

1

Find the values of $a$ and $b$ such that the function $f(x)$ is continuous at $x=2$: $f(x) = \begin{cases} ax + 5 & x < 2 \\ 15 & x = 2 \\ bx^2 - 1 & x > 2 \end{cases}$

[3 marks]

Solution

For continuity at $x = 2$, we require $\text{LHL} = f(2) = \text{RHL}$.

Step 1 — Evaluate LHL and equate to f(2):
$$\text{LHL} = \lim_{x \to 2^-} (ax+5) = 2a+5$$
Set LHL = 15:
$$2a+5 = 15 \Rightarrow 2a = 10 \Rightarrow a = 5$$

Step 2 — Evaluate RHL and equate to f(2):
$$\text{RHL} = \lim_{x \to 2^+} (bx^2-1) = 4b-1$$
Set RHL = 15:
$$4b-1 = 15 \Rightarrow 4b = 16 \Rightarrow b = 4$$

Answer: $a = 5$, $b = 4$.

∴ a = 5, b = 4

One-Page Revision Cheat Sheet

1

L-R-F check is mandatory: A function is continuous at a point $x=a$ if and only if LHL = RHL = Functional Value.

2

Neighborhood bounds: A standard neighborhood of $a$ contains $a$, while a deleted neighborhood excludes $a$.

3

Limit existence boundary: If LHL and RHL approach different finite values, the limit does not exist, creating a jump discontinuity.

4

The Absolute Value rule: For $|x-a|$ terms, replace with $-(x-a)$ when evaluating LHL ($x < a$) and with $+(x-a)$ for RHL ($x > a$).

5

Classification of holes: If the limit exists but $f(a)$ is undefined, the discontinuity is removable and can be solved by redefining $f(a)$.

6

Jump magnitude: The jump size is a non-negative distance calculated as $|\text{RHL} - \text{LHL}|$.

7

Indeterminate form handling: Never leave an answer as $0/0$. Use algebraic factoring, rationalization, or trig identities.

8

Squeeze Theorem trick: Bound functions like $x^2 \cos(1/x)$ between $-x^2$ and $x^2$ to evaluate limits at 0.

9

Piecewise systems: For variables like $a$ and $b$, establish limits at each transition boundary to create solvable equations.