$\lim_{x\to 0}(\cos x +\sin x)^{1/\tan x}$

Question

I have the following limit:

$\lim_{x\to 0}(\cos x +\sin x)^{1/\tan x}$

I did the following:

$1+\frac{1}{f(x)}=\cos x+\sin x$

$f(x)=\frac{1}{\cos x+\sin x -1}$

$\lim_{x\to 0}\left(\left((\cos x +\sin x)^\frac{1}{\cos x+\sin x-1}\right)^\frac{\cos x+\sin x-1}{\tan x}\right)$

$=e^\left(\lim_{x\to 0}\frac{\cos x+\sin x-1}{\tan x}\right)$

This is where I got stuck since I found no way to modify $\frac{\cos x+\sin x-1}{\tan x}$ such that I could find the limit

Answer 1

$$\lim _{ x\rightarrow 0 }{ \frac { \cos { x+\sin { x } -1 } }{ \tan { x } } } \overset { L'hospital }{ = } \lim _{ x\rightarrow 0 }{ \frac { -\sin { x } +\cos { x } }{ \frac { 1 }{ \cos ^{ 2 }{ x } } } } =1$$

Without L'Hospital rule second approach $$\lim _{ x\rightarrow 0 }{ \frac { \cos { x+\sin { x } -1 } }{ \tan { x } } } =\lim _{ x\rightarrow 0 }{ \frac { -2\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) +2\sin { \left( \frac { x }{ 2 } \right) \cos { \left( \frac { x }{ 2 } \right) } } } }{ \frac { 2\sin { \left( \frac { x }{ 2 } \right) \cos { \left( \frac { x }{ 2 } \right) } } }{ \cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } } } } =\\ =\lim _{ x\rightarrow 0 }{ \frac { -\sin { \left( \frac { x }{ 2 } \right) +\cos { \left( \frac { x }{ 2 } \right) } } }{ \cos { \left( \frac { x }{ 2 } \right) } } \left[ \cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } \right] } =1$$

Answer 2

taking log , we get

$$\frac {1}{\tan (x)}\ln (\cos (x)+\sin (x)) $$

now use

$$\ln (1+X)\sim X \;\;(X\to 0) $$

and $$\tan (X)\sim X $$

the limit becomes

$$\lim_0 \frac {\cos (x)+\sin (x)-1}{x} $$ $$=1$$ YOUR LIMIT IS THEN $$e $$

Answer 3

$$\lim_{x\to 0}(\cos x +\sin x)^{1/\tan x}=\lim_{x\to 0}(\cos x)^{1/\tan x}\cdot\lim_{x\to 0}(1+\tan x)^{1/\tan x}.$$

For the first factor,

$$(\cos x)^{1/\tan x}=((1-\sin^2x)^{1/\sin^2x})^{\sin x\cos x/2}$$

which tends to $(e^{-1})^0=1.$

The second factor is obviously $e$.

Answer 4

From $\;\lim\limits_{x\to0}\left(1+x\right)^{1/x}=e\;$ , we get that (this is not trivial, it requires proof...but it isn't too hard):

$$\lim_{x\to x_0}(1+f(x))^{1/f(x)}=e\;,\;\;\text{whenever}\;\;\lim_{x\to x_0}f(x)=0$$

and thus

$$\lim_{x\to0}(\cos x+\sin x)^{1/\tan x}=\lim_{x\to0}\cos^{1/\tan x}\cdot\left(1+\tan x\right)^{1/\tan x}=1\cdot e=e$$