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  • Evaluate: <img alt="\mathrm{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x}}.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow%200

Evaluate: \mathrm{\lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x}}.}
 

Option: 1

-1


Option: 2

0


Option: 3

-\frac{1}{2}


Option: 4

1


Answers (1)

best_answer

Here, the form is \mathrm{1^{\infty}.}

\mathrm{\text { Limit } =e^{\lim _{x \rightarrow 0} \log \left(\frac{\tan x}{x}\right)^{\frac{1}{x}}} }

\mathrm{=e^{\lim _{x \rightarrow 0} \frac{1}{x} \log \left(\frac{\tan x}{x}\right)}}.........(1)

Now, \mathrm{ \lim _{x \rightarrow 0} \frac{1}{x} \log \frac{\tan x}{x}=\lim _{x \rightarrow 0} \frac{\log \frac{\tan x}{x}}{x} \quad}       \mathrm{ \left ( From:\frac{0}{0} \right )}

\mathrm{ =\lim _{x \rightarrow 0} \frac{\frac{x}{\tan x} \cdot \frac{x \sec ^2 x-\tan x}{x^2}}{1} }              (using L' Hospital's rule)

\mathrm{ =\lim _{x \rightarrow 0} \frac{x \sec ^2 x-\tan x}{x \tan x} \quad }                      \mathrm{\left ( From :\frac{0}{0} \right )}

\mathrm{ =\lim _{x \rightarrow 0} \frac{\sec ^2 x+x \cdot 2 \sec ^2 x \tan x-\sec ^2 x}{\tan x+x \sec ^2 x} }(using L' Hospital's rule)

\mathrm{ =\lim _{x \rightarrow 0} \frac{\sec ^2 x+x \cdot 2 \sec ^2 x \tan x-\sec ^2 x}{\tan x+x \sec ^2 x} }        \mathrm{ \left ( From:\frac{0}{0} \right ) }

\mathrm{ =\lim _{x \rightarrow 0} \frac{2 \sec ^2 x \cdot \tan x+2 x\left(\sec ^4 x+2 \sec ^2 x \cdot \tan ^2 x\right)}{2 \sec ^2 x+x \cdot 2 \sec ^2 x \tan x} }

\mathrm{ =\frac{0}{2}=0 }

\mathrm{\therefore from (1), limit =e^0=1.}

Hence option 4 is correct.





 

Posted by

manish painkra

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