Punjab Board Class 11 Mathematics

Chapter 12: Limit and Continuity — Class 11 Maths Notes

Limit and continuity: one-sided limits and the standard algebraic techniques, explained clearly. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 12.

Exercise solutions — Chapter 12

Every exercise of Limit and Continuity solved step by step.

Exercise 12.1 Exercise 12.2 Exercise 12.3

Mathematics 11 (Punjab Board)

Chapter 12: Limit and Continuity — Theory

12.1 The Limit of a Function

We say that the limit of a function \(f(x)\) as \(x\) approaches \(c\) is \(L\), written

\lim_{x \to c} f(x) = L

if the values of \(f(x)\) get closer and closer to \(L\) as \(x\) gets closer and closer to \(c\) from both sides. Is mein \(x = c\) par function ki value ka koi role nahi hota — limit sirf nazdeek ki values par depend karti hai.

Key point: The limit describes behaviour near \(c\), not at \(c\). A function can have a limit at a point where it is not even defined.

One-Sided Limits

The left-hand limit \(\lim_{x \to c^-} f(x)\) uses values of \(x\) slightly less than \(c\); the right-hand limit \(\lim_{x \to c^+} f(x)\) uses values slightly greater than \(c\). The (two-sided) limit exists exactly when both one-sided limits exist and are equal:

\lim_{x \to c} f(x) = L \quad \Longleftrightarrow \quad \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L

Piecewise functions ke liye hamesha dono sides alag alag check karo — yehi exam ka favourite sawal hai.

Key point: If LHL \(\neq\) RHL, the limit does not exist.

12.1 Theorems on Limits of Functions

If \(\lim_{x \to c} f(x) = L\) and \(\lim_{x \to c} g(x) = M\) both exist, then:

\lim_{x \to c} [f(x) \pm g(x)] = L \pm M

\lim_{x \to c} [f(x) \cdot g(x)] = L \cdot M, \qquad \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M} \quad (M \neq 0)

\lim_{x \to c} [k f(x)] = kL, \qquad \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}

Polynomial aur rational functions mein seedha value put kar do — bas denominator zero nahi hona chahiye. Agar \(\frac{0}{0}\) form ban jaye to algebraic technique (factorize ya rationalize) lagao.

Key point: Direct substitution works unless it gives \(\frac{0}{0}\) or an infinite form — then simplify first.

Limits at Infinity and Limits of Sequences

For a rational function or sequence, divide numerator and denominator by the highest power of the variable, then use \(\frac{1}{n} \to 0\) as \(n \to \infty\):

\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} \quad (\text{equal degrees})

Agar numerator ki degree zyada ho to limit \(\pm \infty\) (divergent); agar denominator ki degree zyada ho to limit \(0\). For square roots with \(x \to -\infty\), remember \(\sqrt{x^2} = |x| = -x\).

Key point: Equal degrees → ratio of leading coefficients; denominator heavier → 0; numerator heavier → infinite.

12.2 Limits of Important Functions

These standard limits are used again and again — inko yaad rakho:

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

\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1, \qquad \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0

Trigonometric limits mein pehle expression ko \(\frac{\sin \theta}{\theta}\) ki form mein convert karo — multiply/divide ya substitution se. \(\theta\) hamesha radians mein hota hai.

Key point: \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) only works when \(\theta \to 0\) in radians and the angle matches the denominator exactly.

Limits Involving the Number \(e\)

The number \(e \approx 2.718\) is defined through limits:

\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e, \qquad \lim_{x \to 0} (1 + x)^{1/x} = e

General form: \(\lim (1 + \text{small})^{\text{big}} = e^{\text{(small × big)}}\). Masalan \(\lim_{n \to \infty} (1 + \frac{3}{n})^{2n} = e^{3 \times 2} = e^6\). One-sided exponential limits bhi yaad rakho: \(x \to 0^+\) par \(e^{1/x} \to \infty\), aur \(x \to 0^-\) par \(e^{1/x} \to 0\).

Key point: Rewrite the base so the fraction is \(\frac{1}{\text{exponent}}\) — the power of \(e\) is (inside coefficient) × (outside exponent).

12.3 Continuity and Discontinuity of Functions

A function \(f\) is continuous at \(x = c\) if all three conditions hold:

\text{(i) } f(c) \text{ is defined}, \qquad \text{(ii) } \lim_{x \to c} f(x) \text{ exists}, \qquad \text{(iii) } \lim_{x \to c} f(x) = f(c)

Agar teeno mein se koi ek bhi fail ho jaye to function \(x = c\) par discontinuous hai. Common cases: \(f(c)\) defined nahi (hole), LHL \(\neq\) RHL (jump), ya limit \(\neq f(c)\) (removable jump).

Key point: Check the three conditions in order — most students lose marks by skipping \(f(c)\).

12.4 Applications of Transcendental Functions

Exponential models describe growth and decay. For decay \(A(t) = A_0 e^{-kt}\) with \(k > 0\):

\lim_{t \to \infty} A_0 e^{-kt} = 0

For growth \(P(t) = P_0 e^{rt}\) with \(r > 0\), the limit as \(t \to \infty\) is \(\infty\). Logistic-type models like \(P(t) = \frac{K}{1 + ae^{-bt}}\) approach the carrying capacity \(K\) as \(t \to \infty\), because \(e^{-bt} \to 0\).

Key point: As \(t \to \infty\): \(e^{-kt} \to 0\) (decay) and \(e^{rt} \to \infty\) (growth) — the sign of the exponent decides everything.

Mathematics 11 (Punjab Board)

Solved Examples: Chapter 12 – Limit and Continuity

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 12: Limit and Continuity, worked in full.

Example 1: Evaluate (i) \(\lim_{x \to 1} \frac{x^2 - 1}{x^2 - x}\) (ii) \(\lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}}\)

Direct substitution se \(\frac{0}{0}\) form ban rahi hai, is liye pehle factorize ya rationalize karenge. (i) mein common factor cancel karenge, (ii) mein conjugate se multiply karenge.

(i)

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

(ii)

\lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} \frac{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})}{\sqrt{x} - \sqrt{3}} = \lim_{x \to 3} (\sqrt{x} + \sqrt{3}) = 2\sqrt{3}

Answer: (i) \(2\), (ii) \(2\sqrt{3}\)

Example 2: Find the limit of the sequence \(a_n = \frac{2n + 3}{n + 1}\) as \(n \to \infty\)

Sequence ki limit ke liye numerator aur denominator ko \(n\) ki highest power se divide karenge, phir \(\frac{1}{n} \to 0\) use karenge.

\lim_{n \to \infty} \frac{2n + 3}{n + 1} = \lim_{n \to \infty} \frac{2 + \frac{3}{n}}{1 + \frac{1}{n}} = \frac{2 + 0}{1 + 0} = 2

Answer: \(2\)

Example 3: Evaluate \(\lim_{x \to +\infty} \frac{5x^4 - 10x^2 + 1}{-3x^3 + 10x^2 + 50}\)

Numerator aur denominator ko \(x\) ki highest power (\(x^4\)) se divide karenge. Degree ka farq dekh kar sign ka khayal rakhenge.

\lim_{x \to +\infty} \frac{5x^4 - 10x^2 + 1}{-3x^3 + 10x^2 + 50} = \lim_{x \to +\infty} \frac{5 - \frac{10}{x^2} + \frac{1}{x^4}}{-\frac{3}{x} + \frac{10}{x^2} + \frac{50}{x^4}} = \frac{5 - 0 + 0}{-0 + 0 + 0}

Numerator \(5\) ki taraf jata hai jabke denominator negative side se \(0\) ki taraf jata hai, is liye quotient \(-\infty\) ki taraf diverge hota hai.

= -\infty

Answer: \(-\infty\) (the limit diverges to negative infinity). Note: some printed keys list \(\infty\); the leading coefficients \(5\) and \(-3\) have opposite signs, so the correct divergence is \(-\infty\).

Example 4: Evaluate \(\lim_{x \to -\infty} \frac{4x^4 - 5x^3}{-3x^5 + 2x^2 + 1}\)

Highest power \(x^5\) se divide karenge. Denominator ki degree zyada hai, is liye limit \(0\) hogi.

\lim_{x \to -\infty} \frac{4x^4 - 5x^3}{-3x^5 + 2x^2 + 1} = \lim_{x \to -\infty} \frac{\frac{4}{x} - \frac{5}{x^2}}{-3 + \frac{2}{x^3} + \frac{1}{x^5}} = \frac{0 - 0}{-3 + 0 + 0} = 0

Answer: \(0\)

Example 5: Evaluate (i) \(\lim_{x \to -\infty} \frac{2 - 3x}{\sqrt{3 + 4x^2}}\) (ii) \(\lim_{x \to +\infty} \frac{2 - 3x}{\sqrt{3 + 4x^2}}\)

Square root ke sath infinity limit mein \(\sqrt{x^2} = |x|\) use hota hai. \(x \to -\infty\) par \(|x| = -x\), aur \(x \to +\infty\) par \(|x| = x\).

(i) \(x \to -\infty\), so divide by \(-x\):

\lim_{x \to -\infty} \frac{2 - 3x}{\sqrt{3 + 4x^2}} = \lim_{x \to -\infty} \frac{\frac{2}{-x} + 3}{\sqrt{\frac{3}{x^2} + 4}} = \frac{0 + 3}{\sqrt{0 + 4}} = \frac{3}{2}

(ii) \(x \to +\infty\), so divide by \(x\):

\lim_{x \to +\infty} \frac{2 - 3x}{\sqrt{3 + 4x^2}} = \lim_{x \to +\infty} \frac{\frac{2}{x} - 3}{\sqrt{\frac{3}{x^2} + 4}} = \frac{0 - 3}{\sqrt{0 + 4}} = -\frac{3}{2}

Answer: (i) \(\frac{3}{2}\), (ii) \(-\frac{3}{2}\)

Common mistake: Dividing by \(x\) instead of \(-x\) when \(x \to -\infty\) flips the sign of the answer. Always use \(|x|\).

Example 6: Express each limit in terms of \(e\): (i) \(\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{2n}\) (ii) \(\lim_{n \to 0} (1 + 2n)^{1/n}\)

Standard form \(\lim_{m \to \infty} (1 + \frac{1}{m})^m = e\) banane ke liye power ko adjust karenge. Power of \(e\) = (andar ka coefficient) × (bahar ka exponent).

(i)

\left(1 + \frac{3}{n}\right)^{2n} = \left[\left(1 + \frac{3}{n}\right)^{\frac{n}{3}}\right]^6 \quad \Rightarrow \quad \lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{2n} = e^6

(ii) Put \(m = 2n\); as \(n \to 0\), \(m \to 0\):

\lim_{n \to 0} (1 + 2n)^{1/n} = \lim_{m \to 0} \left[(1 + m)^{1/m}\right]^2 = e^2

Answer: (i) \(e^6\), (ii) \(e^2\)

Example 7: Evaluate \(\lim_{\theta \to 0} \frac{\sin 7\theta}{\theta}\)

Standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) use karenge. Angle aur denominator match karne ke liye \(7\) se multiply/divide karenge.

\lim_{\theta \to 0} \frac{\sin 7\theta}{\theta} = \lim_{\theta \to 0} \frac{\sin 7\theta}{7\theta} \times 7 = 1 \times 7 = 7

Answer: \(7\)

Example 8: Evaluate \(\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta}\)

Conjugate \((1 + \cos \theta)\) se multiply aur divide karenge taake \(\sin^2 \theta = 1 - \cos^2 \theta\) ban jaye, phir standard limit lagayenge.

\frac{1 - \cos \theta}{\theta} \cdot \frac{1 + \cos \theta}{1 + \cos \theta} = \frac{1 - \cos^2 \theta}{\theta(1 + \cos \theta)} = \frac{\sin^2 \theta}{\theta(1 + \cos \theta)} = \sin \theta \cdot \frac{\sin \theta}{\theta} \cdot \frac{1}{1 + \cos \theta}

\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0 \times 1 \times \frac{1}{2} = 0

Answer: \(0\)

Example 9: Determine whether \(\lim_{x \to 2} f(x)\) and \(\lim_{x \to 4} f(x)\) exist, where \(f(x) = \begin{cases} 2x + 1 & \text{if } 0 \le x \le 2 \\ 7 - x & \text{if } 2 < x \le 4 \\ x & \text{if } x > 4 \end{cases}\)

Piecewise function mein har point par left-hand aur right-hand limits alag nikalte hain. Dono barabar hon to limit exist karti hai.

(i) At \(x = 2\):

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 1) = 5, \qquad \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (7 - x) = 5

Since LHL = RHL = 5, \(\lim_{x \to 2} f(x)\) exists and equals \(5\).

(ii) At \(x = 4\):

\lim_{x \to 4^-} f(x) = \lim_{x \to 4^-} (7 - x) = 3, \qquad \lim_{x \to 4^+} f(x) = \lim_{x \to 4^+} (x) = 4

Since LHL \(\neq\) RHL, \(\lim_{x \to 4} f(x)\) does not exist.

Answer: \(\lim_{x \to 2} f(x) = 5\) exists; \(\lim_{x \to 4} f(x)\) does not exist.

Example 10: Discuss the continuity of \(f(x) = \frac{x^2 - 1}{x - 1}\) at \(x = 1\)

Continuity ki pehli shart hai ke \(f(c)\) defined ho. Yahan denominator zero ho jata hai, is liye pehli shart hi fail ho jayegi.

f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0} \quad \text{(undefined)}

Since \(f(1)\) is not defined, condition (i) of continuity fails.

Answer: \(f\) is discontinuous at \(x = 1\) (a hole in the graph).

Example 11: Discuss the continuity of \(f(x) = 3x^2 - 5x + 4\) at \(x = 1\)

Polynomial functions har point par continuous hoti hain. Teenon conditions check karenge: \(f(1)\), limit, aur unki equality.

\lim_{x \to 1} f(x) = \lim_{x \to 1} (3x^2 - 5x + 4) = 3 - 5 + 4 = 2, \qquad f(1) = 3 - 5 + 4 = 2

Since \(\lim_{x \to 1} f(x) = f(1) = 2\), all three conditions hold.

Answer: \(f\) is continuous at \(x = 1\).

Example 12: Discuss the continuity at \(x = 3\) of (a) \(f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{if } x \neq 3 \\ 6 & \text{if } x = 3 \end{cases}\) (b) \(g(x) = \frac{x^2 - 9}{x - 3}\)

Dono functions mein limit to \(6\) hai, lekin farq \(x = 3\) par defined hone ka hai. Wahi continuity ka faisla karega.

(a)

\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x + 3)(x - 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6 = f(3)

All three conditions hold, so \(f\) is continuous at \(x = 3\) — the hole has been filled.

(b)

\(g(3)\) is not defined (division by zero), so condition (i) fails.

Answer: (a) \(f\) is continuous at \(x = 3\); (b) \(g\) is discontinuous at \(x = 3\).

Example 13: Discuss the continuity of \(f(x) = \begin{cases} x - 1 & \text{if } x < 3 \\ 2x + 1 & \text{if } x \ge 3 \end{cases}\) at \(x = 3\)

Pehle \(f(3)\) nikalenge, phir LHL aur RHL. Agar LHL \(\neq\) RHL to limit exist nahi karegi aur function discontinuous hoga.

\[f(3) = 2(3) + 1 = 7\]

\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x - 1) = 2, \qquad \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x + 1) = 7

Since LHL \(\neq\) RHL, \(\lim_{x \to 3} f(x)\) does not exist — condition (ii) fails.

Answer: \(f\) is discontinuous at \(x = 3\) (jump discontinuity).

Example 14: Radioactive decay is given by \(A(t) = A_0 e^{-kt}\) (\(k > 0\)). Find the limit of the amount as \(t \to \infty\).

Decay model mein exponent negative hai. \(t \to \infty\) par \(e^{-kt} \to 0\), is liye poora amount zero ki taraf jata hai.

\lim_{t \to \infty} A(t) = \lim_{t \to \infty} A_0 e^{-kt} = A_0 \cdot 0 = 0

Answer: \(0\) — eventually the substance decays completely.

Example 15: An investment grows by continuous compounding as \(A(t) = P_0 e^{rt}\) (\(r > 0\)). What happens to its value as \(t \to \infty\)?

Growth model mein exponent positive hai, is liye \(t \to \infty\) par \(e^{rt} \to \infty\).

\lim_{t \to \infty} A(t) = \lim_{t \to \infty} P_0 e^{rt} = P_0 \cdot \infty = \infty

Answer: \(\infty\) — the investment grows without bound.

Example 16: The demand function is \(D(p) = \frac{100}{p + 1}\). Find \(\lim_{p \to \infty} D(p)\).

Price bohat zyada barhne par denominator infinity ki taraf jata hai, to fraction zero ki taraf jata hai.

\lim_{p \to \infty} D(p) = \lim_{p \to \infty} \frac{100}{p + 1} = 0

Answer: \(0\) — at very high prices, demand vanishes.

Example 17: A star's apparent brightness follows \(B(d) = \frac{L}{d^2}\). Find \(\lim_{d \to \infty} B(d)\).

Inverse square law: distance infinity ki taraf jaye to brightness zero ki taraf jati hai.

\lim_{d \to \infty} B(d) = \lim_{d \to \infty} \frac{L}{d^2} = 0

Answer: \(0\) — a star infinitely far away appears dark.

Mathematics 11 (Punjab Board)

MCQs: Chapter 12 – Limit and Continuity

\(\lim_{x \to 2} (3x - 1) =\)
1. \(5\)
2. \(6\)
3. \(4\)
4. \(7\)
Direct substitution: \(3(2) - 1 = 5\). Polynomial mein seedha value put karo.
\(\lim_{x \to 1} \frac{x^2 - 1}{x - 1} =\)
1. \(0\)
2. \(1\)
3. \(2\)
4. does not exist
Factorize: \(\frac{(x-1)(x+1)}{x-1} = x + 1 \to 2\). Pehle \((x-1)\) cancel karo.
\(\lim_{x \to 0} \frac{\sin 3x}{x} =\)
1. \(0\)
2. \(1\)
3. \(3\)
4. \(\frac{1}{3}\)
\(\frac{\sin 3x}{3x} \times 3 \to 1 \times 3 = 3\). Angle aur denominator match karo.
\(\lim_{n \to \infty} \frac{3n^2 + 1}{2n^2 - n} =\)
1. \(\frac{3}{2}\)
2. \(3\)
3. \(0\)
4. \(\infty\)
Equal degrees: ratio of leading coefficients \(= \frac{3}{2}\).
\(\lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^x =\)
1. \(e\)
2. \(e^2\)
3. \(1\)
4. \(0\)
Standard form \((1 + \frac{2}{x})^x \to e^2\) — power of \(e\) is (inside coefficient) × (outside exponent) \(= 2 \times 1\).
If \(\lim_{x \to c} f(x) = 4\) and \(\lim_{x \to c} g(x) = -1\), then \(\lim_{x \to c} [f(x) \cdot g(x)] =\)
1. \(-4\)
2. \(4\)
3. \(3\)
4. \(-3\)
Product theorem of limits: \(4 \times (-1) = -4\).
\(\lim_{x \to 0} \frac{1 - \cos x}{x} =\)
1. \(0\)
2. \(1\)
3. \(\frac{1}{2}\)
4. does not exist
Conjugate se \(\frac{\sin^2 x}{x(1+\cos x)} = \sin x \cdot \frac{\sin x}{x} \cdot \frac{1}{1+\cos x} \to 0\).
\(f(x) = \frac{x^2 - 4}{x - 2}\) (\(x \neq 2\)) is discontinuous at \(x = 2\) because
1. \(f(2)\) is undefined
2. the limit does not exist
3. the limit is not equal to \(f(2)\)
4. none of these
Continuity ki pehli shart hi fail: \(x = 2\) par denominator zero, to \(f(2)\) defined nahi.
\(\lim_{x \to -\infty} \frac{2x}{x + 1} =\)
1. \(2\)
2. \(-2\)
3. \(0\)
4. \(\infty\)
\(x\) se divide karo: \(\frac{2}{1 + 1/x} \to 2\). Sign ka farq nahi parta jab degrees equal hon.
\(\lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h} =\)
1. \(\frac{1}{4}\)
2. \(\frac{1}{2}\)
3. \(4\)
4. \(0\)
Rationalize: \(\frac{1}{\sqrt{4+h}+2} \to \frac{1}{4}\).
\(\lim_{x \to 0} \frac{e^{2x} - 1}{x} =\)
1. \(1\)
2. \(2\)
3. \(e\)
4. \(0\)
Standard limit \(\frac{e^{kx}-1}{x} \to k\), yahan \(k = 2\).
Which of the following is continuous at \(x = 0\)?
1. \(f(x) = \frac{1}{x}\)
2. \(f(x) = |x|\)
3. \(f(x) = \frac{x^2}{x}\) (\(x \neq 0\))
4. \(f(x) = \frac{1}{x^2}\)
\(|x|\) har point par defined aur continuous hai; baqi \(x = 0\) par defined hi nahi.
\(\lim_{x \to 3} \frac{x^2 - 9}{x - 3} =\)
1. \(3\)
2. \(6\)
3. \(0\)
4. does not exist
\(\frac{(x-3)(x+3)}{x-3} = x + 3 \to 6\).
\(\lim_{x \to \pi/2} \sin x =\)
1. \(0\)
2. \(1\)
3. \(-1\)
4. does not exist
\(\sin\) continuous hai, to \(\sin(\pi/2) = 1\).
If \(f\) is continuous at \(x = c\), then
1. \(\lim_{x \to c} f(x) = f(c)\)
2. \(f(c)\) is undefined
3. LHL \(\neq\) RHL
4. none of these
Yahi continuity ki definition hai — teenon shartein isi mein summarize hain.
\(\lim_{n \to \infty} \frac{n^3}{n^2 + 1} =\)
1. \(0\)
2. \(1\)
3. \(\infty\)
4. \(\frac{1}{2}\)
Numerator ki degree zyada hai, is liye sequence diverge karti hai.
\(\lim_{x \to 0} \frac{\tan x}{x} =\)
1. \(0\)
2. \(1\)
3. \(\infty\)
4. \(-1\)
\(\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \to 1 \times 1 = 1\).
\(\lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} =\)
1. \(-1\)
2. \(0\)
3. \(1\)
4. does not exist
\(x \to 0^+\) par \(e^{1/x} \to \infty\); \(e^{1/x}\) se divide karne par \(\frac{1-0}{1+0} = 1\).
\(\lim_{\theta \to 0} \frac{\sin 5\theta}{\sin 2\theta} =\)
1. \(\frac{5}{2}\)
2. \(\frac{2}{5}\)
3. \(1\)
4. \(0\)
\(\frac{5\theta}{2\theta} \cdot \frac{\sin 5\theta}{5\theta} \cdot \frac{2\theta}{\sin 2\theta} \to \frac{5}{2}\).
A function has a jump discontinuity at \(x = c\) if
1. LHL \(\neq\) RHL
2. \(f(c)\) is undefined
3. the limit is infinite
4. none of these
Jump discontinuity ka matlab graph mein uchhal — dono sides ki limits mukhtalif hoti hain.
\(\lim_{x \to 2} \frac{|x - 2|}{x - 2} =\)
1. \(1\)
2. \(-1\)
3. \(0\)
4. does not exist
LHL \(= -1\) aur RHL \(= 1\); barabar nahi, is liye limit exist nahi karti.
\(\lim_{x \to \infty} \frac{2x^2 - 3x + 1}{5x^2 + 2} =\)
1. \(\frac{2}{5}\)
2. \(2\)
3. \(0\)
4. \(\infty\)
Equal degrees: leading coefficients ka ratio \(= \frac{2}{5}\).

Mathematics 11 (Punjab Board)

Chapter 12: Limit and Continuity — Formula Sheet

The Limit and One-Sided Limits

\[\lim_{x \to c} f(x) = L\]

\[\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L\]

Theorems on Limits of Functions

\[\lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)\]

\[\lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)\]

\[\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} \quad \left(\lim_{x \to c} g(x) \neq 0\right)\]

\[\lim_{x \to c} [k f(x)] = k \lim_{x \to c} f(x), \qquad \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to c} f(x)}\]

Limits at Infinity

\[\lim_{n \to \infty} \frac{1}{n} = 0, \qquad \lim_{n \to \infty} \frac{1}{n^k} = 0 \quad (k > 0)\]

\[\sqrt{x^2} = |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}\]

Limits of Important Functions

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

\[\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1, \qquad \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = 0\]

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

Limits Involving \(e\)

\[\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e, \qquad \lim_{x \to 0} (1 + x)^{1/x} = e\]

\[\lim_{x \to 0^+} e^{1/x} = \infty, \qquad \lim_{x \to 0^-} e^{1/x} = 0\]

Continuity and Discontinuity

\[f \text{ continuous at } c \iff \begin{cases} \text{(i) } f(c) \text{ is defined} \\ \text{(ii) } \lim_{x \to c} f(x) \text{ exists} \\ \text{(iii) } \lim_{x \to c} f(x) = f(c) \end{cases}\]

Transcendental (Exponential) Models

\[\lim_{t \to \infty} A_0 e^{-kt} = 0 \quad (k > 0), \qquad \lim_{t \to \infty} P_0 e^{rt} = \infty \quad (r > 0)\]

\[\lim_{t \to \infty} \frac{K}{1 + ae^{-bt}} = K \quad (b > 0)\]

Chapter 12 FAQs

What is covered in Chapter 12 (Limit and Continuity)?

This chapter covers Limits, one-sided limits, continuity, discontinuity, algebraic techniques. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.

How many marks does Chapter 12 carry in the Punjab Board paper?

Chapter 12 (Limit and Continuity) carries approximately 13 marks in the Punjab Board Class 11 Mathematics paper.

Are these notes for the new PCTB syllabus?

Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.

Where are the exercise solutions for Chapter 12?

Use the exercise index on this page — every exercise of Chapter 12 has its own page with complete step-by-step solutions.