Limit Continuity and Differentiability 1 Question 2
3. If $\lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}=\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}$, then $k$ is
(a) $\frac{4}{3}$
(b) $\frac{3}{8}$
(c) $\frac{3}{2}$
(2019 Main, 10 April I)
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Answer:
Correct Answer: 3. (d)
Solution:
Key Idea $\lim _{x \rightarrow a} f(x)$ exist iff
$\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x)$
At $x=0$,
$\begin{aligned} \text { RHL } & =\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}} \ & =\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x-\sin (x \cdot 0))^{2}}{x^{2}}\end{aligned}$
$\because|x|=x$ for $x>0$ and $[x]=0$ for $0<x<1$
$$ \begin{aligned} & =\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}}{x^{2}} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \cdot \frac{\pi \sin ^{2} x}{x^{2}}+1 \end{aligned} $$
$$ \begin{aligned} & =\pi \lim _{x \rightarrow 0^{+}} \frac{\tan \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \cdot \lim _{x \rightarrow 0^{+}} \frac{\sin ^{2} x}{x^{2}}+1 \\ & =\pi+1 \quad \because \quad \lim _{x \rightarrow 0} \frac{\tan x}{x}=1 \\ & \quad \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \end{aligned} $$
and LHL
$$ \begin{aligned} & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+\left(|x|-\sin (x[x])^{2}\right.}{x^{2}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+\left(-x-\sin (x(-1))^{2}\right.}{x^{2}} \\ & \because|x|=-x \text { for } x<0 \\ & \text { and }[x]=-1 \text { for }-1<x<0 \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x+\sin (-x))^{2}}{x^{2}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+(x-\sin x)^{2}}{x^{2}} \\ & {[\because \sin (-\theta)=-\sin \theta]} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)+x^{2}+\sin ^{2} x-2 x \sin x}{x^{2}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)}{x^{2}}+1+\frac{\sin ^{2} x}{x^{2}}-\frac{2 x \sin x}{x^{2}} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \cdot \frac{\pi \sin ^{2} x}{x^{2}}+1+\frac{\sin ^{2} x}{x^{2}}-2 \frac{\sin x}{x} \\ & =\lim _{x \rightarrow 0^{-}} \frac{\tan \left(\pi \sin ^{2} x\right)}{\pi \sin ^{2} x} \cdot \lim _{x \rightarrow 0^{-}} \frac{\pi \sin ^{2} x}{x^{2}}+ \\ & 1+\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} x}{x^{2}}-2 \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x} \\ & =\pi+1+1-2=\pi \end{aligned} $$
$\because$ RHL $\neq$ LHL
$\therefore$ Limit does not exist
