Question

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

Answer 1

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

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,

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

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

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

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

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

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

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

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

$=e^{ab}$

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