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  • Find the limit of the following:<img alt="\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow%200%7D%5C

Find the limit of the following:\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}

Option: 1

1


Option: 2

0


Option: 3

\infty


Option: 4

e


Answers (1)

best_answer

Given: \lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}

If \lim _{x \rightarrow a} f(x)=1 and \lim _{x \rightarrow a} g(x)=\infty, then \lim _{x \rightarrow a} f(x)^{g\left ( x \right )}= e^{\lim _{x \rightarrow a} \left ( f\left ( x \right ) -1\right )\left ( g\left ( x \right ) \right )}.

=\lim _{x \rightarrow 0}\left[1+\left(\frac{\sin x}{x}-1\right)\right]^{1 / x}
=e^{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}-1\right) \frac{1}{x}}
=e^{\lim _{x \rightarrow 0}\left(\frac{\sin x-x}{x^{2}}\right) }

Applying L-Hospital’s rule, the above expression becomes,

=e^{\lim _{x \rightarrow 0}\left(\frac{\cos x-1}{2 x}\right)}
=e^{\frac{-1}{2} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x}\right)}

Using the property of limit \lim _{x \rightarrow 0} \frac{1-\cos x}{x}=0,then the expression becomes,

=e^0
=1

Therefore,\lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]^{1 / x}= 1

Posted by

Divya Prakash Singh

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