Question

Evaluate the limit:
$\underset{x\to 0}{lim}(\cos x+\sin x{)}^{1/x}$

Answer 1

Given:
$\underset{x\to 0}{lim}(\cos x+\sin x{)}^{1/x}$

It becomes $1^{\infty }$ form

Using the result below:

If $\underset{x\to a}{lim}f\left(x\right)=1$ and $\underset{x\to a}{lim}g\left(x\right)=\infty$ such that $\underset{x\to a}{lim}\left\{f\right(x)-1\}g\left(x\right)$ exists, then

$\underset{x\to a}{lim}\left\{f\right(x){\}}^{g\left(x\right)}=e^{\underset{x\to a}{lim}\left\{f\right(x)-1\}g\left(x\right)}$

Here,

$f\left(x\right)=\cos x+\sin x$

$g\left(x\right)=\frac{1}{x}$

$\underset{x\to 0}{lim}(\cos x+\sin x{)}^{1/x}$

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

$=e^{\underset{x\to 0}{lim}(\frac{\sin x}{x}-\frac{(1-\cos x)}{x})}$

$=e^{\underset{x\to 0}{lim}(\frac{\sin x}{x}-\frac{2{\sin }^{2\frac{x}{2}}}{x})}$

$[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=e^{\underset{x\to 0}{lim}(\frac{\sin x}{x}-\frac{2\sin \left(\frac{x}{2}\right)\times \sin \left(\frac{x}{2}\right)}{2\times \frac{x}{2}})}$

$=e^{\underset{x\to 0}{lim}(\frac{\sin x}{x}-\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\times \sin \left(\frac{x}{2}\right))}$

$=e^{1-1\times 0}$
$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1]$

$=e^1=e$

Therefore,
$\underset{x\to 0}{lim}(\cos x+\sin x{)}^{1/x}=e$