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Punjab Board Class 11 Mathematics

Exercise 12.1 Solutions — Limit and Continuity

Complete worked solutions for Exercise 12.1 from Chapter 12 (Limit and Continuity), Punjab Board Class 11 Mathematics. Each question shows every step and names the formula used.

Mathematics 11 (Punjab Board)

Exercise 12.1 Solutions

Worked solutions for Punjab Board Class 11 Mathematics, Chapter 12: Limit and Continuity.

Question 2: Find the limit of the following sequences, if it exists: (i) \(a_n = \frac{2n+3}{n+1}\) (ii) \(b_n = \frac{2n+3}{n^2+1}\) (iii) \(c_n = \frac{5n^2}{2n+3}\) (iv) \(d_n = \frac{n^2-3n+1}{2n^2+n+4}\)

Sequence ki limit ke liye numerator aur denominator ko \(n\) ki highest power se divide karenge, phir \(\frac{1}{n} \to 0\) use karenge. Degrees compare karne se pehle hi andaza lag jata hai: equal degrees → leading coefficients ka ratio.

(i) Divide by \(n\):

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

(ii) Divide by \(n^2\) (denominator heavier → 0):

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

(iii) Divide by \(n\) (numerator heavier → infinite):

\[\lim_{n \to \infty} \frac{5n^2}{2n+3} = \lim_{n \to \infty} \frac{5n}{2 + \frac{3}{n}} = \frac{\infty}{2} = \infty \quad \text{(divergent)}\]

(iv) Divide by \(n^2\):

\[\lim_{n \to \infty} \frac{n^2-3n+1}{2n^2+n+4} = \lim_{n \to \infty} \frac{1 - \frac{3}{n} + \frac{1}{n^2}}{2 + \frac{1}{n} + \frac{4}{n^2}} = \frac{1 - 0 + 0}{2 + 0 + 0} = \frac{1}{2}\]
Answer: (i) \(2\), (ii) \(0\), (iii) \(\infty\) (divergent), (iv) \(\frac{1}{2}\)

Question 3: Evaluate each limit by using theorems of limits: (i) \(\lim_{x \to 3} (2x+4)\) (ii) \(\lim_{x \to 1} (3x^2-2x+4)\) (iii) \(\lim_{x \to 3} \sqrt{x^2+x+4}\) (iv) \(\lim_{x \to 2} \sqrt{x^2+4}\) (v) \(\lim_{x \to 2} (\sqrt{x^2+1} - \sqrt{x^2+5})\) (vi) \(\lim_{x \to 2} \frac{2x^3+5x}{3x-2}\)

Yahan direct substitution kaam karegi kyunke koi \(\frac{0}{0}\) form nahi ban rahi. Sum, difference, product, quotient aur root theorems of limits use honge. (Part (iii) mein key ke mutabiq \(x^2\) use ho raha hai.)

(i)

\[\lim_{x \to 3} (2x+4) = 2(3) + 4 = 10\]

(ii)

\[\lim_{x \to 1} (3x^2-2x+4) = 3(1)^2 - 2(1) + 4 = 5\]

(iii)

\[\lim_{x \to 3} \sqrt{x^2+x+4} = \sqrt{9 + 3 + 4} = \sqrt{16} = 4\]

(iv)

\[\lim_{x \to 2} \sqrt{x^2+4} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\]

(v)

\[\lim_{x \to 2} (\sqrt{x^2+1} - \sqrt{x^2+5}) = \sqrt{5} - \sqrt{9} = \sqrt{5} - 3\]

(vi)

\[\lim_{x \to 2} \frac{2x^3+5x}{3x-2} = \frac{2(8) + 10}{6 - 2} = \frac{26}{4} = \frac{13}{2}\]
Answer: (i) \(10\), (ii) \(5\), (iii) \(4\), (iv) \(2\sqrt{2}\), (v) \(\sqrt{5} - 3\), (vi) \(\frac{13}{2}\)

Question 4: Evaluate each limit by using algebraic techniques: (i) \(\lim_{x \to -1} \frac{x^3-x}{x+1}\) (ii) \(\lim_{x \to 3} \frac{x^2-5x+6}{x^2-2x-3}\) (iii) \(\lim_{x \to 2} \frac{x^3-8}{x^2-5x+6}\) (iv) \(\lim_{x \to 1} \frac{x^3-3x^2+3x-1}{x^3-x}\) (v) \(\lim_{x \to 2} \frac{x^3-6x^2+12x-8}{x^3-4x}\) (vi) \(\lim_{x \to 1} \frac{x^4-1}{x^3-3x+2}\) (vii) \(\lim_{x \to 2} \frac{x-2}{\sqrt{x+2}-\sqrt{6-x}}\) (viii) \(\lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\) (ix) \(\lim_{x \to a} \frac{x^n-a^n}{x-a}\)

Direct value put karne par \(\frac{0}{0}\) form ban rahi hai, is liye factorize ya rationalize karke common factor cancel karenge. Parts (ii) aur (iii) mein key ke mutabiq corrected limits (\(x \to 3\) aur \(x \to 2\)) use ho rahi hain.

(i) Factorize and cancel \((x+1)\):

\[\lim_{x \to -1} \frac{x^3-x}{x+1} = \lim_{x \to -1} \frac{x(x-1)(x+1)}{x+1} = \lim_{x \to -1} x(x-1) = (-1)(-2) = 2\]

(ii) Factorize and cancel \((x-3)\):

\[\lim_{x \to 3} \frac{x^2-5x+6}{x^2-2x-3} = \lim_{x \to 3} \frac{(x-2)(x-3)}{(x-3)(x+1)} = \lim_{x \to 3} \frac{x-2}{x+1} = \frac{1}{4}\]

(iii) Use \(x^3 - 8 = (x-2)(x^2+2x+4)\) and cancel \((x-2)\):

\[\lim_{x \to 2} \frac{x^3-8}{x^2-5x+6} = \lim_{x \to 2} \frac{(x-2)(x^2+2x+4)}{(x-2)(x-3)} = \lim_{x \to 2} \frac{x^2+2x+4}{x-3} = \frac{12}{-1} = -12\]

(iv) Use \((x-1)^3\) and cancel one factor \((x-1)\):

\[\lim_{x \to 1} \frac{(x-1)^3}{x(x-1)(x+1)} = \lim_{x \to 1} \frac{(x-1)^2}{x(x+1)} = \frac{0}{2} = 0\]

(v) Use \((x-2)^3\) and cancel \((x-2)\):

\[\lim_{x \to 2} \frac{(x-2)^3}{x(x-2)(x+2)} = \lim_{x \to 2} \frac{(x-2)^2}{x(x+2)} = \frac{0}{8} = 0\]

(vi) Factorize: numerator \((x-1)(x+1)(x^2+1)\); denominator \((x-1)^2(x+2)\):

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

At \(x = 1\) the numerator is \(4\) but the denominator is still \(0\) — the function is unbounded near \(x = 1\) (tending to \(+\infty\) from the right and \(-\infty\) from the left), so the two-sided limit does not exist.

(vii) Rationalize the denominator with its conjugate:

\[\lim_{x \to 2} \frac{x-2}{\sqrt{x+2}-\sqrt{6-x}} \cdot \frac{\sqrt{x+2}+\sqrt{6-x}}{\sqrt{x+2}+\sqrt{6-x}} = \lim_{x \to 2} \frac{(x-2)(\sqrt{x+2}+\sqrt{6-x})}{2x-4}\]
\[= \lim_{x \to 2} \frac{\sqrt{x+2}+\sqrt{6-x}}{2} = \frac{2+2}{2} = 2\]

(viii) Rationalize the numerator:

\[\lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} = \frac{1}{2\sqrt{x}}\]

(ix) Standard limit formula:

\[\lim_{x \to a} \frac{x^n-a^n}{x-a} = na^{n-1}\]
Answer: (i) \(2\), (ii) \(\frac{1}{4}\), (iii) \(-12\), (iv) \(0\), (v) \(0\), (vi) does not exist (unbounded), (vii) \(2\), (viii) \(\frac{1}{2\sqrt{x}}\), (ix) \(na^{n-1}\). Note: some printed keys list \(0\) for part (vi); the function is unbounded near \(x = 1\), so the correct conclusion is that the limit does not exist.

Common mistake: Cancelling \((x - a)\) and then substituting \(x = a\) into a denominator that still has \((x - a)\) left over — part (vi) shows why you must check what remains.

Question 5: Evaluate the following limits: (i) \(\lim_{x \to 0} \frac{\sin 5x}{x}\) (ii) \(\lim_{x \to 0} \frac{\sin x^2}{x}\) (iii) \(\lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta}\) (iv) \(\lim_{x \to \pi/4} \frac{\sin x - \cos x}{x - \pi/4}\) (v) \(\lim_{x \to 0} \frac{\cos ax - \cos bx}{x^2}\) (vi) \(\lim_{x \to \pi/4} \frac{\tan x - 1}{x - \pi/4}\) (vii) \(\lim_{x \to 0} \frac{1-\cos 2x}{x^2}\) (viii) \(\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\) (ix) \(\lim_{x \to 1} \frac{x^3-1}{x^2-1}\) (x) \(\lim_{x \to 3} \frac{x^2 - x\log x + 3\log x - 9}{x-3}\) (xi) \(\lim_{x \to 0} \frac{x(2^x-1)}{1-\cos x}\)

Trigonometric limits mein standard result \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) aur identities (\(\cos A - \cos B\), \(1 - \cos \theta = 2\sin^2 \frac{\theta}{2}\)) use karenge. Angle aur denominator ko match karna zaroori hai.

(i) Multiply and divide by \(5\):

\[\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \times 5 = 5\]

(ii) Write as \(\frac{\sin x^2}{x^2} \times x\):

\[\lim_{x \to 0} \frac{\sin x^2}{x} = \lim_{x \to 0} \frac{\sin x^2}{x^2} \times x = 1 \times 0 = 0\]

(iii) Use half-angle identities:

\[\lim_{\theta \to 0} \frac{1-\cos \theta}{\sin \theta} = \lim_{\theta \to 0} \frac{2\sin^2 \frac{\theta}{2}}{2\sin \frac{\theta}{2}\cos \frac{\theta}{2}} = \lim_{\theta \to 0} \tan \frac{\theta}{2} = 0\]

(iv) Substitute \(h = x - \frac{\pi}{4}\):

\[\lim_{h \to 0} \frac{\sin(\frac{\pi}{4}+h) - \cos(\frac{\pi}{4}+h)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{2}}(\cos h + \sin h) - \frac{1}{\sqrt{2}}(\cos h - \sin h)}{h}\]
\[= \lim_{h \to 0} \frac{\sqrt{2}\sin h}{h} = \sqrt{2}\]

(v) Use \(\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\):

\[\lim_{x \to 0} \frac{-2\sin\frac{a+b}{2}x \sin\frac{a-b}{2}x}{x^2} = -2 \cdot \frac{a+b}{2} \cdot \frac{a-b}{2} = \frac{b^2-a^2}{2}\]

(vi) Substitute \(h = x - \frac{\pi}{4}\) and use \(\tan(\frac{\pi}{4}+h) = \frac{1+\tan h}{1-\tan h}\):

\[\lim_{h \to 0} \frac{\frac{1+\tan h}{1-\tan h} - 1}{h} = \lim_{h \to 0} \frac{2\tan h}{h(1-\tan h)} = 2 \times 1 \times 1 = 2\]

(vii) Use \(1 - \cos 2x = 2\sin^2 x\):

\[\lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2\left(\lim_{x \to 0} \frac{\sin x}{x}\right)^2 = 2\]

(viii) Using the result of part (v) for numerator and denominator:

\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx} = \frac{\frac{b^2-a^2}{2}}{\frac{d^2-c^2}{2}} = \frac{b^2-a^2}{d^2-c^2}\]

(ix) Factorize and cancel \((x-1)\):

\[\lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+1)} = \frac{3}{2}\]

(x) Group terms to expose the factor \((x-3)\):

\[\lim_{x \to 3} \frac{(x^2-9) - \log x\,(x-3)}{x-3} = \lim_{x \to 3} \frac{(x-3)(x+3-\log x)}{x-3} = 6 - \log 3\]

(xi) Use \(1 - \cos x = 2\sin^2 \frac{x}{2}\) and \(\lim_{x \to 0} \frac{2^x-1}{x} = \ln 2\):

\[\lim_{x \to 0} \frac{x(2^x-1)}{2\sin^2 \frac{x}{2}} = \left(\lim_{x \to 0} \frac{2^x-1}{x}\right) \cdot \frac{x^2}{2\sin^2 \frac{x}{2}} = \ln 2 \times 2 \times 1 = 2\ln 2\]
Answer: (i) \(5\), (ii) \(0\), (iii) \(0\), (iv) \(\sqrt{2}\), (v) \(\frac{b^2-a^2}{2}\), (vi) \(2\), (vii) \(2\), (viii) \(\frac{b^2-a^2}{d^2-c^2}\), (ix) \(\frac{3}{2}\), (x) \(6 - \log 3\), (xi) \(2\ln 2\)

Question 6: Express each limit in terms of \(e\): (i) \(\lim_{n \to \infty} (1+\frac{1}{n})^{2n}\) (ii) \(\lim_{n \to \infty} (1+\frac{1}{n})^{n/2}\) (iii) \(\lim_{n \to \infty} (1-\frac{1}{n})^n\) (iv) \(\lim_{n \to \infty} (1+\frac{1}{3n})^n\) (v) \(\lim_{x \to \infty} (1+\frac{4}{x})^x\) (vi) \(\lim_{x \to 0} (1+3x)^{2/x}\) (vii) \(\lim_{x \to 0} (1+2x^2)^{1/x^2}\) (viii) \(\lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x}\) (ix) \(\lim_{x \to 0} (1-x)^{1/x}\) (x) \(\lim_{x \to 0^-} \frac{e^{1/x}-1}{e^{1/x}+1}\) (xi) \(\lim_{x \to 0^+} \frac{e^{1/x}-1}{e^{1/x}+1}\) (xii) \(\lim_{x \to 2} \frac{e^x-e^2}{x-2}\)

Standard forms \(\lim_{n \to \infty} (1+\frac{1}{n})^n = e\) aur \(\lim_{x \to 0} (1+x)^{1/x} = e\) use karenge. Base ko is form mein convert karke exponent adjust karenge. (Part (v) mein key ke mutabiq \(x \to \infty\) use ho raha hai.)

(i)

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

(ii)

\[\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^{n/2} = \left[\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n\right]^{1/2} = e^{1/2} = \sqrt{e}\]

(iii)

\[\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n = \lim_{n \to \infty} \left(1+\frac{-1}{n}\right)^n = e^{-1} = \frac{1}{e}\]

(iv)

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

(v)

\[\lim_{x \to \infty} \left(1+\frac{4}{x}\right)^x = e^4\]

(vi)

\[\lim_{x \to 0} (1+3x)^{2/x} = \lim_{x \to 0} \left[(1+3x)^{1/(3x)}\right]^6 = e^6\]

(vii)

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

(viii) Use \(\lim_{x \to 0} \frac{e^{kx}-1}{x} = k\):

\[\lim_{x \to 0} \frac{e^{ax}-e^{bx}}{x} = \lim_{x \to 0} \left(\frac{e^{ax}-1}{x} - \frac{e^{bx}-1}{x}\right) = a - b\]

(ix)

\[\lim_{x \to 0} (1-x)^{1/x} = \lim_{x \to 0} (1+(-1)x)^{1/x} = e^{-1} = \frac{1}{e}\]

(x) As \(x \to 0^-\), \(\frac{1}{x} \to -\infty\) so \(e^{1/x} \to 0\):

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

(xi) As \(x \to 0^+\), \(\frac{1}{x} \to \infty\) so \(e^{1/x} \to \infty\); divide top and bottom by \(e^{1/x}\):

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

(xii) This is the derivative of \(e^x\) at \(x = 2\):

\[\lim_{x \to 2} \frac{e^x-e^2}{x-2} = e^2\]
Answer: (i) \(e^2\), (ii) \(\sqrt{e}\), (iii) \(\frac{1}{e}\), (iv) \(e^{1/3}\), (v) \(e^4\), (vi) \(e^6\), (vii) \(e^2\), (viii) \(a-b\), (ix) \(\frac{1}{e}\), (x) \(-1\), (xi) \(1\), (xii) \(e^2\)

Common mistake: In parts (x) and (xi), students treat \(0^+\) and \(0^-\) the same — the behaviour of \(e^{1/x}\) is completely different on each side.