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  • How to calculate   <img alt="\mathrm{ \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

How to calculate   \mathrm{ \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} }

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

-1


Answers (1)

By taking cases \mathrm{x \leq 0}  and \mathrm{x \geq 0} we obtain the limits

                 \mathrm{ |\tan x| \geq|\sin x| \Longrightarrow\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} \geq 1 }
and

                \mathrm{ \left\{\begin{array}{l} \sec x+\tan x \geq 1+\tan x \\ \cos x+\tan x \leq 1+\tan x \end{array} \Longrightarrow\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{\sin x}} \leq(\sec x)^{\frac{1}{|\sin x|}}\right. }

which means with your substitution we have

                   \mathrm{ \lim _{t \rightarrow 0} 1 \leq L \leq \lim _{t \rightarrow 0}\left(1-t^2\right)^{-\frac{1}{2|| \mid}} }

The limit on the left is 1 and the limit on the right is

                     \mathrm{ \lim _{t \rightarrow 0}\left(1-t^2\right)^{-\frac{|c|}{2 t^2}}=\lim _{t \rightarrow 0}\left(\left(1-t^2\right)^{\frac{1}{2}}\right)^{-\frac{|| t \mid}{2}}=\left(e^{-1}\right)^0=1 }

Thus the limit we want is 1 by squeeze theorem.

Posted by

Sumit Saini

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