SATHEE: Limit Continuity and Differentiability 2 Question 5

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, $lim_{x \to 0^{-}} \frac{x ([x] + | x |) \sin [x]}{| x |}$ is equal to

(a) 0

(b) $\sin 1$

(d) 1

(2019 Main, 9 Jan II)

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Answer:

Correct Answer: 5. (a)

Solution:

We have,

$\begin{aligned} Y_{n} = \frac{1}{n} [(n + 1) (n + 2) \dots (n + n)]^{1 / n} \\ and lim_{n \to \infty} Y_{n} = L \\ \Rightarrow L = lim_{n \to \infty} \frac{1}{n} [(n + 1) (n + 2) (n + 3) \dots (n + n)]^{1 / n} \\ \Rightarrow L = lim_{n \to \infty} 1 + \frac{1}{n} 1 + \frac{2}{n} 1 + \frac{3}{n} \dots 1 + \frac{n}{n} \frac{1}{n} \\ \Rightarrow \log L = lim_{n \to \infty} \frac{1}{n} \log 1 + \frac{1}{n} + \log 1 + \frac{2}{n} \dots \log 1 + \frac{n}{n} \\ \Rightarrow \log L = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \log 1 + \frac{r}{n} \\ \Rightarrow \log L = \int_{0}^{1} 1 \times \log_{I} (1 + x) d x \\ \Rightarrow \log L = (x \cdot \log (1 + x))_{0}^{1} - \int_{0}^{1} \frac{d}{d x} (\log (1 + x) \int d x d x \end{aligned}$

$\Rightarrow \log L = [x \log (1 + x)]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x} d x$

[by using integration by parts]

$\Rightarrow \log L = \log 2 - \int_{0}^{1} \frac{x + 1}{x + 1} - \frac{1}{x + 1} d x$

$\Rightarrow \log L = \log 2 - [x]_{0}^{1} + [\log (x + l)]_{0}^{1}$

$\Rightarrow \log L = \log 2 - 1 + \log 2 - 0$

$\Rightarrow \log L = \log 4 - \log e = \log \frac{4}{e} \Rightarrow L = \frac{4}{e} \Rightarrow$

$[L] = \frac{4}{e} = 1$

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, $lim_{x \to 0^{-}} \frac{x ([x] + | x |) \sin [x]}{| x |}$ is equal to

Answer:

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Limit Continuity and Differentiability 2 Question 5

5. For each x∈R, let [x] be the greatest integer less than or equal to x. Then, limx→0−x([x]+|x|)sin⁡[x]|x| is equal to

Answer:

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5. For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then, $lim_{x \to 0^{-}} \frac{x ([x] + | x |) \sin [x]}{| x |}$ is equal to