For x>0,limx→0 (sin⁡x)1/x+1xsin⁡x is equal to

Solution:

Here ,

$\begin{array}{r} lim_{x \to 0} (\sin x)^{1 / x} + lim_{x \to 0} {(\frac{1}{x})}^{\sin x} = 0 + lim_{x \to 0} e^{\log {(\frac{1}{x})}^{\sin x}} \\ [∵ lim_{x \to 0} (\sin x)^{1 / x} \to 0 as 0 < \sin x < 1] \end{array}$

$= e^{x \to 0} \frac{\log (1 / x)}{cosec x} = e^{x \to 0} \frac{x (- \frac{1}{x^{2}})}{- cosec xcot x}$ [using L' Hospital's rule]

$= e^{lim_{x \to 0} \frac{\sin x}{x} \tan x} = e^{0} = 1$

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