Limit Continuity and Differentiability 2 Question 5
5. For each $x \in R$, let $[x]$ be the greatest integer less than or equal to $x$. Then, $\lim _{x \rightarrow 0^{-}} \frac{x([x]+|x|) \sin [x]}{|x|}$ is equal to
(a) 0
(b) $\sin 1$
(c) $-\sin 1$
(d) 1
(2019 Main, 9 Jan II)
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Answer:
Correct Answer: 5. (a)
Solution:
- We have,
$$ \begin{aligned} & Y _n=\frac{1}{n}[(n+1)(n+2) \ldots(n+n)]^{1 / n} \\ & \text { and } \lim _{n \rightarrow \infty} Y _n=L \\ & \Rightarrow \quad L=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2)(n+3) \ldots(n+n)]^{1 / n} \\ & \Rightarrow \quad L=\lim _{n \rightarrow \infty} 1+\frac{1}{n} \quad 1+\frac{2}{n} \quad 1+\frac{3}{n} \ldots 1+\frac{n}{n} \frac{1}{n} \\ & \Rightarrow \log L=\lim _{n \rightarrow \infty} \frac{1}{n} \log 1+\frac{1}{n}+\log 1+\frac{2}{n} \ldots \log 1+\frac{n}{n} \\ & \Rightarrow \log L=\lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=1}^{n} \log 1+\frac{r}{n} \\ & \Rightarrow \log L=\int _0^{1} 1 \times \log _I(1+x) d x \\ & \Rightarrow \log L=(x \cdot \log (1+x)) _0^{1}-\int _0^{1} \frac{d}{d x}\left(\log (1+x) \int d x d x\right. \end{aligned} $$
$\Rightarrow \log L=[x \log (1+x)] _0^{1}-\int _0^{1} \frac{x}{1+x} d x$
[by using integration by parts]
$\Rightarrow \log L=\log 2-\int _0^{1} \frac{x+1}{x+1}-\frac{1}{x+1} dx$
$\Rightarrow \log L=\log 2-[x] _0^{1}+[\log (x+l)] _0^{1}$
$\Rightarrow \log L=\log 2-1+\log 2-0$
$\Rightarrow \log L=\log 4-\log e=\log \frac{4}{e} \Rightarrow L=\frac{4}{e} \Rightarrow$
$[L]=\frac{4}{e}=1$
