Q49 / 120 · JEE Advanced 2006 · Mathematics

For $x>0, \lim _{x \rightarrow 0}\left((\sin x)^{1 / x}+\left(\frac{1}{x}\right)^{\sin x}\right)$ is

  • A
    0
  • B
    $-1$
  • Correct answer
    C
    1
  • D
    2

Solution

Here, $\lim _{x \rightarrow 0}(\sin x)^{1 / x}+\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x} \Rightarrow 0+\lim _{x \rightarrow 0} e^{\log \left(\frac{1}{x}\right)^{\sin x}} \quad\left\{\begin{array}{l}\lim _{x \rightarrow 0}(\sin x)^{1 / x} \rightarrow 0 \\ \text { as, (decimal) })^x \rightarrow 0\end{array}\right\}$ $\Rightarrow \quad e^{\lim _{x \rightarrow 0} \frac{\log (1 / x)}{\operatorname{cosec} x}}$ On applying L Hospital's Rule, we get $$ \Rightarrow \quad e^{\lim _{x \rightarrow 0} \frac{x\left(-\frac{1}{x^2}\right)}{-\operatorname{cosec} x \cot x}}=e^{\lim _{x \rightarrow 0} \frac{\sin x}{x} \tan x}=e^0=1 $$

Hence, the correct answer is option: 'C'.

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