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  • Evaluate:  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin ^2 x}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightar

Evaluate:  \mathrm{\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin ^2 x}}

Option: 1

0


Option: 2

2


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

                             \mathrm{ \lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin ^2 x}=L }

Using L'hopital:

                             \mathrm{\lim _{x \rightarrow 0} \frac{\frac{\sec ^2 x}{2 \sqrt{1+\tan x}}-\frac{1}{2 \sqrt{1+x}}}{2 \sin x \cos x}=L}

Reordering the denominator:

                             \mathrm{\lim _{x \rightarrow 0} \frac{\frac{\sec ^2 x}{2 \sqrt{1+\tan x}}-\frac{1}{2 \sqrt{1+x}}}{\sin 2 x}=L}

Using L'hopital Again 

                            \mathrm{\lim _{x \rightarrow 0} \frac{\frac{-2 \sec ^2 x \tan x \sqrt{1+\tan x}-\frac{\sec ^4 x}{2 \sqrt{1+\tan x}}}{2(1+\tan x)}+\frac{1}{4 \sqrt{1+x^3}}}{2 \cos 2 x}=L}

You cannot use l'hopital again because  \mathrm{\lim _{x \rightarrow 0} \cos 2 x=1>0}
so replacing:
                     \mathrm{ \lim _{x \rightarrow 0} \frac{\frac{-2 \sec ^2 x \tan x \sqrt{1+\tan x}-\frac{\sec ^4 x}{2 \sqrt{1+\tan x}}}{2(1+\tan x)}+\frac{1}{4 \sqrt{1+x^3}}}{2 \cos 2 x}=\frac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2 \cdot 1}=0 }

Posted by

shivangi.bhatnagar

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