Question

For x > 0, $ lim_{ x \to 0} \Bigg [ (sin \, x)^{1/x} + \bigg( \frac{1}{x}\bigg)^{sin \, x} \Bigg ] $ is

• A. 0
• B. -2
• C. 1
• D. 2

Answer 1

The Correct Option is C

Here, $ lim_{ x \ to 0} (sin \ x)^{1/x} + lim_{x \to 0} \bigg(\frac{1}{x}\bigg)^{sin \ x} $
= 0 + $ lim_{ x \to 0} e^{log \bigg(\frac{1}{x}\bigg)^{sin \ x} } = \ e^{lim_{ x \to 0} \frac{log \ (1 / x)}{cosec \ x}}$ $\hspace21mm$ $\Bigg [ \begin{array}
\ lim_{ x \to 0} (sin \ x)^{1/x} \rightarrow 0 \\
as, (decimal)^\infty \ \ \rightarrow 0
\end{array} \Bigg ] $
Appling L'Hospital's rule, we get
$e^lim_{x \to 0 } \frac{ x \bigg( - \frac{1}{x^2}\bigg)}{ - cosec \ x \ cot \ x} = e^{lim_{x \to 0 } \frac{sin \ x}{ x} tan \ x} = e^0 = 1 $