1. ## Fantastic limit

a real paramtre and

Calculate :

2. Originally Posted by dhiab
a real paramtre and

Calculate :
$\lim_{x\to a}\left( \frac{\tan x- \tan a}{x-a}-\frac{\sin x-\sin a}{x-a} \right) \frac{x-a}{x-a-(\sin x-\sin a)}$

= $\lim_{x\to a}\left( \frac{\tan x- \tan a}{x-a}-\frac{\sin x-\sin a}{x-a} \right) \frac{1}{1-\frac{\sin x-\sin a}{x-a}}$

$
=\frac{\sec^2a-\cos a}{1-\cos a}
$

$\left(since, \lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)\right)$

You may simplify further but I am bored.

Alternatively you may use the L'Hospital Rule

3. Originally Posted by pankaj
$\lim_{x\to a}\left( \frac{\tan x- \tan a}{x-a}-\frac{\sin x-\sin a}{x-a} \right) \frac{x-a}{x-a-(\sin x-\sin a)}$

= $\lim_{x\to a}\left( \frac{\tan x- \tan a}{x-a}-\frac{\sin x-\sin a}{x-a} \right) \frac{1}{1-\frac{\sin x-\sin a}{x-a}}$

$
=\frac{\sec^2a-\cos a}{1-\cos a}
$

$\left(since, \lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)\right)$

You may simplify further but I am bored.

Alternatively you may use the L'Hospital Rule
Hello : Thank you, I'have anathor resolution with the L'Hospital Rule