JEE Mathematics Limits Indeterminate Forms MCQs Set 04

Answer: (d) \( e^{-2} \)

Explanation:
1. This is a \( 1^\infty \) form. \( \tan\left(\frac\pi4-x\right)=\frac{1-\tan x}{1+\tan x} \), so the base minus 1 is \( \frac{-2\tan x}{1+\tan x} \).
2. \( \frac1x\cdot\frac{-2\tan x}{1+\tan x}\to-2 \), so the limit is \( e^{-2} \).

Teacher's Note:
1. Write tan of a difference.

Answer: (c) \( e^{-2} \)

Explanation:
1. \( \frac{x-1}{x+1}=1-\frac2{x+1} \), a \( 1^\infty \) form.
2. The exponent of e is \( \lim(x+2)\cdot\frac{-2}{x+1}=-2 \), so the limit is \( e^{-2} \).

Teacher's Note:
1. Write the base as 1 plus a small term.

Answer: (c) \( \lambda=1 \) and \( \mu \) is any real constant

Explanation:
1. The limit is \( e^{\lim2x\left(\frac\lambda x+\frac\mu{x^2}\right)}=e^{2\lambda} \).
2. \( e^{2\lambda}=e^2 \) gives \( \lambda=1 \), and \( \mu \) does not affect the limit.

Teacher's Note:
1. Use the 1 to the infinity formula.

Answer: (d) none of these

Explanation:
1. For \( x\to0^+ \), \( \{x\}=x \) and the ratio is \( \frac x{\tan x}\to1 \).
2. For \( x\to0^- \), \( \{x\}=1+x\to1 \) and the ratio tends to \( \frac1{\tan1}\ne1 \).
3. The left and right limits differ, so the limit does not exist.

Teacher's Note:
1. Find the fractional part on each side of 0.

Answer: (a) \( 0 \)

Explanation:
1. For large x, \( x-1\lt [x]\le x \), so \( \ln[x]\le\ln x \).
2. \( \frac{\ln[x]}x\le\frac{\ln x}x\to0 \), and the expression is positive for \( x\ge2 \), so the limit is 0.

Teacher's Note:
1. Bound the greatest integer function.

Answer: (c) not existent

Explanation:
1. For \( x\to1^+ \), \( x-[x]=x-1 \) and the ratio is \( x\cdot\frac{\sin(x-1)}{x-1}\to1 \).
2. For \( x\to1^- \), \( x-[x]\to1 \), so the numerator tends to \( \sin1\ne0 \) while \( x-1\to0^- \), and the ratio tends to \( -\infty \).
3. The limit does not exist.

Teacher's Note:
1. Treat each side of 1 separately.

Answer: (c) not existent

Explanation:
1. For \( x\to0^+ \), \( [x]=0 \) and the ratio is 0.
2. For \( x\to0^- \), \( [x]=-1 \) and \( |x|=-x \): \( \frac{-x}{\sin(-x)}=\frac x{\sin x}\to1 \).
3. The two one sided limits differ, so the limit does not exist.

Teacher's Note:
1. Find [x] on each side of 0.

Answer: (c) does not exist

Explanation:
1. For \( x\to-1^+ \): \( [x]=-1 \), \( |x|\to1 \), the sum tends to 0.
2. For \( x\to-1^- \): \( [x]=-2 \), \( |x|\to1 \), the sum tends to \( -1 \).
3. The one sided limits differ, so the limit does not exist.

Teacher's Note:
1. Compare the two one sided limits.

Answer: (d) none of these

Explanation:
1. For \( x\to0^+ \), \( [x]=0 \), so \( f(x)=0 \).
2. For \( x\to0^- \), \( [x]=-1 \), so \( f(x)=\frac{\sin(-1)}{-1}=\sin1\approx0.84 \).
3. The limits differ, so the limit does not exist.

Teacher's Note:
1. Evaluate [x] on each side.

Answer: (c) \( \frac x2 \)

Explanation:
1. Write \( [kx]=kx-\{kx\} \) with \( 0\le\{kx\}\lt 1 \).
2. The sum is \( x\cdot\frac{n(n+1)}2-\sum\{kx\} \), and \( \sum\{kx\}\lt n \).
3. Dividing by \( n^2 \): the limit is \( \frac x2 \).

Teacher's Note:
1. Use the fractional part to bound the sum.

Answer: (a) \( 1 \)

Explanation:
1. Both numerator and denominator tend to 0. Apply L'Hospital's rule.
2. The derivative of the numerator is \( 2x\cos x^4 \); the denominator's is \( \sin x+x\cos x \).
3. \( \frac{2x\cos x^4}{\sin x+x\cos x}\to\frac{2}{1+1}=1 \).

Teacher's Note:
1. Differentiate the integral using the Leibniz rule.

Answer: (a) \( 0 \)

Explanation:
1. For \( x>1 \), \( \int_1^x(t-1)\,dt=\frac{(x-1)^2}2 \).
2. \( \frac{(x-1)^2/2}{\sin(x-1)}\approx\frac{x-1}2\to0 \).

Teacher's Note:
1. Evaluate the integral on the interval to the right of 1.

Answer: (a) \( 1 \)

Explanation:
1. \( \frac n{\sqrt{9n^2+1}}=\frac1{\sqrt{9+1/n^2}}\to\frac13 \).
2. By continuity the limit is \( f\left(\frac13\right)=1 \).

Teacher's Note:
1. Move the limit inside a continuous function.

Answer: (b) \( -1 \) and \( 2 \)

Explanation:
1. The left hand limit is \( 1+\frac2a \) and the right hand limit is \( a \).
2. \( 1+\frac2a=a\Rightarrow a^2-a-2=0\Rightarrow a=2 \) or \( a=-1 \).

Teacher's Note:
1. Equate the left and right limits.

Answer: (b) \( a \)

Explanation:
1. A repeated root gives \( ax^2+bx+c=a(x-\alpha)^2 \).
2. \( \frac{\sin(a(x-\alpha)^2)}{(x-\alpha)^2}=a\cdot\frac{\sin u}u\to a \) with \( u=a(x-\alpha)^2 \).

Teacher's Note:
1. Write the quadratic as a perfect square.

Answer: (c) \( a=1,\ b=1 \)

Explanation:
1. Dividing, \( \frac{1+n^2}{1+n}=n-1+\frac2{1+n} \).
2. The expression is \( (a-1)n+1-\frac2{1+n} \). For a finite limit the coefficient of n must vanish, so \( a=1 \).
3. Then the limit is \( b=1 \).

Teacher's Note:
1. Divide the polynomials to separate the terms in n.

Answer: (a) A and C

Explanation:
1. By Cramer's rule, \( x=\frac{\alpha\cos\alpha-\sin\alpha}{\sin\alpha-\alpha} \) and \( y=\frac{\tan\alpha-\alpha^2\operatorname{cosec}\alpha}{\sin\alpha-\alpha} \).
2. As \( \alpha\to0 \): numerator of x is \( -\frac{\alpha^3}3 \) and the denominator is \( -\frac{\alpha^3}6 \), so \( x\to2 \).
3. Numerator of y is \( \frac{\alpha^3}6 \), so \( y\to-1 \), i.e. \( y+1=0 \).
4. P tends to \( (2,-1) \), lying on x = 2 and y + 1 = 0.

Teacher's Note:
1. Solve the two lines by Cramer's rule.
2. Expand each trigonometric function as a series.

Answer: (c) \( e^2 \)

Explanation:
1. This is a \( 1^\infty \) form: the limit is \( \exp\left(\lim\frac{f(1+x)-f(1)}{x\,f(1)}\right)=\exp\left(\frac{f'(1)}{f(1)}\right) \).
2. \( \frac{f'(1)}{f(1)}=\frac63=2 \), so the limit is \( e^2 \).

Teacher's Note:
1. Use the formula e raised to the limit of (base minus 1) times the exponent.
2. The difference quotient gives f'(1).

Answer: (d) 5

Explanation:
1. For the limit to be finite, \( f(1+h)\to0 \), so \( f(1)=0 \) (continuity).
2. Then \( f'(1)=\lim\frac{f(1+h)-f(1)}{h}=\lim\frac{f(1+h)}{h}=5 \).

Teacher's Note:
1. A finite limit with h in the denominator forces the numerator to tend to zero.
2. Then recognise the definition of the derivative.

Answer: (a) 8

Explanation:
1. Let \( t=2^x \) (so \( t\to4 \)). Numerator \( =t+\frac8t-6=\frac{(t-2)(t-4)}t \). Denominator \( =\frac1{\sqrt t}-\frac2t=\frac{\sqrt t-2}{t} \).
2. The ratio is \( \frac{(t-2)(t-4)}{\sqrt t-2}=(t-2)(\sqrt t+2) \) since \( t-4=(\sqrt t-2)(\sqrt t+2) \).
3. As \( t\to4 \): \( 2\times4=8 \).

Teacher's Note:
1. Substitute t = 2 to the power x and factor.
2. Cancel the zero factor t - 4 using a difference of squares.