JEE Mathematics Limits Indeterminate Forms MCQs Set 03

Answer: (a) \( \frac12 \)

Explanation:
1. Split the fraction: \( \frac{\sqrt{x-2}}{\sqrt{(x-2)(x+2)}}=\frac1{\sqrt{x+2}}\to\frac12 \).
2. \( \frac{\sqrt x-\sqrt2}{\sqrt{x^2-4}}=\frac{\sqrt{x-2}}{(\sqrt x+\sqrt2)\sqrt{x+2}}\to0 \).
3. The limit is \( \frac12 \).

Teacher's Note:
1. Split the numerator into two parts.
2. Factorise x squared - 4.

Answer: (c) \( \frac12 \)

Explanation:
1. Multiply and divide by the conjugate: \( \frac{\sqrt x}{\sqrt{x+\sqrt x}+\sqrt x} \).
2. Divide by \( \sqrt x \): \( \frac1{\sqrt{1+1/\sqrt x}+1}\to\frac12 \).

Teacher's Note:
1. Rationalise using the conjugate.

Answer: (a) \( \log_e9 \)

Explanation:
1. Multiply by the conjugate: \( \frac{(3^x-1)(\sqrt{x+1}+1)}{x} \).
2. \( \frac{3^x-1}x\to\ln3 \) and \( \sqrt{x+1}+1\to2 \), so the limit is \( 2\ln3=\ln9 \).

Teacher's Note:
1. Use the standard limit of (a to the x - 1) over x.

Answer: (c) \( \log_34 \)

Explanation:
1. Put \( t=\frac1n\to0 \). The expression is \( \frac{(4^t-1)/t}{(3^t-1)/t}\to\frac{\ln4}{\ln3} \).
2. This equals \( \log_34 \).

Teacher's Note:
1. Use the limit (a to the t - 1) over t = ln a.

Answer: (d) none of these

Explanation:
1. Divide by x: \( \frac{(3^x-2^x)/x}{(4^x-3^x)/x}\to\frac{\ln3-\ln2}{\ln4-\ln3}=\frac{\ln(3/2)}{\ln(4/3)} \).
2. This is about 1.41, which is not 1, -1 or 0.

Teacher's Note:
1. Write each difference as a difference of two standard limits.

Answer: (d) \( -\frac1{48} \)

Explanation:
1. The expression is \( \frac{(8+h)^{-1/3}-8^{-1/3}}h \), the derivative of \( (8+x)^{-1/3} \) at \( x=0 \).
2. \( -\frac13(8)^{-4/3}=-\frac13\cdot\frac1{16}=-\frac1{48} \).

Teacher's Note:
1. Recognise a derivative from first principles.

Answer: (c) \( \frac12 \)

Explanation:
1. \( e^x=1+x+\frac{x^2}2+\frac{x^3}6+\dots \).
2. The numerator is \( \frac{x^2}2-\frac{x^3}6-\dots \), so the limit is \( \frac12 \).

Teacher's Note:
1. Expand e to the x as a series.

Answer: (c) \( \frac12 \)

Explanation:
1. \( \tan2x=2x+\frac{8x^3}3+\dots \) and \( 2\tan x=2x+\frac{2x^3}3+\dots \).
2. The numerator is \( x(2x^3+\dots)=2x^4 \). The denominator is \( (2\sin^2x)^2\approx4x^4 \).
3. The limit is \( \frac12 \).

Teacher's Note:
1. Expand tan 2x and tan x to the x cubed term.

Answer: (b) \( 4 \)

Explanation:
1. The sum tends to \( \frac{1/4}{1-1/2}=\frac12 \).
2. \( (0.2)^{\log_{\sqrt5}(1/2)}=5^{-\log_{\sqrt5}(1/2)}=5^{-2\log_5(1/2)}=\left(\frac12\right)^{-2}=4 \).

Teacher's Note:
1. Sum the geometric series first.
2. Write 0.2 as 5 to the power minus 1.

Answer: (a) \( \frac13 \)

Explanation:
1. The rth term is \( \frac1{(3r-2)(3r+1)}=\frac13\left(\frac1{3r-2}-\frac1{3r+1}\right) \).
2. The series telescopes: \( S_n=\frac13\left(1-\frac1{3n+1}\right)\to\frac13 \).

Teacher's Note:
1. Use partial fractions.

Answer: (c) \( \frac{4^{n+1}-1}{3\cdot2^n} \)

Explanation:
1. \( x^2+2x+3=(x+1)^2+2 \), so \( a=2 \). Also \( b=\frac12 \).
2. \( \sum_{r=0}^n2^r\left(\frac12\right)^{n-r}=\frac1{2^n}\sum_{r=0}^n4^r=\frac{4^{n+1}-1}{3\cdot2^n} \).

Teacher's Note:
1. Complete the square for a.
2. Sum the geometric series in 4.

Answer: (b) \( \frac{n(n+1)}2 \)

Explanation:
1. The numerator is \( \sum_{r=1}^n(x^r-1) \).
2. \( \frac{x^r-1}{x-1}\to r \), so the limit is \( \sum r=\frac{n(n+1)}2 \).

Teacher's Note:
1. Write n as the sum of n ones.

Answer: (b) \( 0 \)

Explanation:
1. \( x-\sin x\approx\frac{x^3}6 \) and \( x+\sin^2x\approx x \).
2. The fraction behaves like \( \frac{x^2}6\to0 \), so the limit is 0.

Teacher's Note:
1. Expand sin x to the cubic term.

Answer: (a) \( -\frac12 \)

Explanation:
1. \( \cos x=1-\frac{x^2}2+\dots \), so \( \ln\cos x\approx-\frac{x^2}2 \).
2. The limit is \( -\frac12 \).

Teacher's Note:
1. Use ln(1 + u) is approximately u.

Answer: (a) \( \frac12\log_2e \)

Explanation:
1. \( 1-\cos x\approx\frac{x^2}2 \) and \( 2^x-1\approx x\ln2 \).
2. The limit is \( \frac1{2\ln2}=\frac12\log_2e \).

Teacher's Note:
1. Use standard series for both factors.

Answer: (b) \( 1 \)

Explanation:
1. Both numerator and denominator tend to 0. Use L'Hospital's rule.
2. The derivative ratio is \( \frac{-f'(x)/(2\sqrt{f(x)})}{-1/(2\sqrt x)} \), which at \( x=4 \) is \( \frac{f'(4)/4}{1/4}=1 \).

Teacher's Note:
1. Differentiate the numerator and the denominator.

Answer: (a) \( 6 \)

Explanation:
1. Write \( f(x)=f(0)+xf'(0)+\frac{x^2}2f''(0)+\dots \).
2. The constant terms: \( 2-3+1=0 \). The x terms: \( 2-6+4=0 \). The \( x^2 \) terms: \( \frac{f''(0)}2(2-12+16)=3f''(0) \).
3. With \( f''(0)=2 \) the limit is 6.

Teacher's Note:
1. Expand each term as a Taylor polynomial.

Answer: (c) \( 2 \)

Explanation:
1. With \( f(1)=g(1)=2 \), the numerator is \( 2g(x)-2f(x)-2+2=2\{g(x)-f(x)\} \).
2. So the ratio equals 2 wherever \( g(x)\ne f(x) \), and the limit is 2.

Teacher's Note:
1. Substitute the given values into the numerator.

Answer: (b) \( e^a \)

Explanation:
1. This is a \( 1^\infty \) form: the limit is \( e^{\lim n\sin(a/n)} \).
2. \( n\sin\frac an=a\cdot\frac{\sin(a/n)}{a/n}\to a \), so the limit is \( e^a \).

Teacher's Note:
1. Use the formula e to the limit of g times (f - 1).

Answer: (d) none of these

Explanation:
1. Taking logarithms, \( \frac1{x^2}\ln\frac{1+5x}{1+3x^2}\approx\frac{5x}{x^2}=\frac5x \).
2. As \( x\to0^+ \) this tends to \( +\infty \) and as \( x\to0^- \) it tends to \( -\infty \), so the limit does not exist (it is \( \infty \) from the right and 0 from the left).

Teacher's Note:
1. Take the logarithm and look at the sign of x.