The value of lim x 0 ( ( sin x ) 1 / x + ( 1 x ) sin x ) , where x 0 is [2006]

Question

The value of $\lim_{x \to 0} ({(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}),$ where $x > 0$ is [2006]

Smart Achievers · Accepted Answer

(3)

$\lim_{x \to 0} [{(\sin x)}^{1 / x} + {(\frac{1}{x})}^{\sin x}]$

$= \lim_{x \to 0} {(\sin x)}^{1 / x} + \lim_{x \to 0} {(\frac{1}{x})}^{\sin x}$

$= 0 + e^{\lim_{x \to 0} \sin x \log (\frac{1}{x})}$ $(∵ | \sin x | < 1 when x \to 0)$

$= e^{\lim_{x \to 0} \frac{- \log x}{cosec x}}$ $= e^{\lim_{x \to 0} \frac{- 1 / x}{- cosec x \cot x}}$ (Using L'Hospital rule)

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

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