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

Answer: (C)

1

Answer 2

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
$e^lim_{x \to 0 } \frac{ x \bigg( - \frac{1}{x^2}\bigg)}{ - cosec \ x \ cot \ x} = e^{lim_{x \to 0 } \frac{sin \ x}{ x} tan \ x} = e^0 = 1$