hello can some please show me how a question of this sort is evaluated.
lim (tan x -sec x) as x tends to pi/2.
thanks.
Correct!
(1)Express the limit in the terms of sin(x) and cos(x).
(2)Use L Hopital's rule.
The correct answer is $\displaystyle 0$.
To check click here.
sbhatnagar suggested you us L'Hopital's rule. Do you know what that is?
If $\displaystyle \lim_{x\to a} f(x)= 0$ and $\displaystyle \lim_{x\to a} g(x)= 0$, then
$\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}$
Note that you differentiate the numerator and denominator separately, you do not differentiate the fraction.
and if you can't use L'Hopital ...
$\displaystyle \lim_{x \to \frac{\pi}{2}} \tan{x} - \sec{x}$
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin{x}-1}{\cos{x}}$
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin{x}-1}{\cos{x}} \cdot \frac{\sin{x}+1}{\sin{x}+1}$
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}(\sin{x}+1)}$
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{-\cos^2{x}}{\cos{x}(\sin{x}+1)}$
$\displaystyle \lim_{x \to \frac{\pi}{2}} \frac{-\cos{x}}{\sin{x}+1} = \frac{0}{2} = 0$