Question

Evaluate the limit:
$\underset{x\to \pi /2}{lim}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}$

Answer 1

Given:
$\underset{x\to \pi /2}{lim}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}$

$=\underset{x\to \pi /2}{lim}\frac{a^{\cos x}(a^{\cot x-\cos x}-1)}{\cot x-\cos x}$

$=a^{\cos \frac{\pi }{2}}\underset{x\to \pi /2}{lim}\frac{(a^{\cot x-\cos x}-1)}{\cot x-\cos x}$

Put $y=\cot \alpha -\cos \alpha$
$[\text{As}x\to \frac{\pi }{2}\Rightarrow \cot x-\cos x=\cot \frac{\pi }{2}-\cos \frac{\pi }{2}=0]$
$\therefore y\to 0$

$=a^0\underset{y\to 0}{lim}\frac{(a^y-1)}{y}$

$=1\times \underset{y\to 0}{lim}\frac{(a^y-1)}{y}$

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

Therefore,
$\underset{x\to \pi /2}{lim}\frac{a^{\cot x}-a^{\cos x}}{\cot x-\cos x}=\log a$