Math Help - Evaluting limit

1. Evaluting limit

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.

2. Re: Evaluting limit

First, realize that it's of the indeterminate form ∞ - ∞ .

Express tan(x) and sec(x) in terms of sin(x) & cos(x) . Combine them into a single rational expression.

See what you then have.

Use L'Hôpital's rule if necessary.

3. Re: Evaluting limit

Correct!

(1)Express the limit in the terms of sin(x) and cos(x).
(2)Use L Hopital's rule.

The correct answer is $0$.

4. Re: Evaluting limit

Originally Posted by SammyS
First, realize that it's of the indeterminate form ∞ - ∞ .

Express tan(x) and sec(x) in terms of sin(x) & cos(x) . Combine them into a single rational expression.

See what you then have.

Use L'Hôpital's rule if necessary.
Ok I have expressed it in terms of sin and cos.

(Sinx-1)/cos x. Should I differenciate? Can you plz show me how to arrive at the answer from here. Thanks.

5. Re: Evaluting limit

sbhatnagar suggested you us L'Hopital's rule. Do you know what that is?

If $\lim_{x\to a} f(x)= 0$ and $\lim_{x\to a} g(x)= 0$, then
$\lim_{x\to a}\frac{f(x)}{g(x)}= \lim_{x\to a}\frac{f'(x)}{g'(x)}$

Note that you differentiate the numerator and denominator separately, you do not differentiate the fraction.

6. Re: Evaluting limit

and if you can't use L'Hopital ...

$\lim_{x \to \frac{\pi}{2}} \tan{x} - \sec{x}$

$\lim_{x \to \frac{\pi}{2}} \frac{\sin{x}-1}{\cos{x}}$

$\lim_{x \to \frac{\pi}{2}} \frac{\sin{x}-1}{\cos{x}} \cdot \frac{\sin{x}+1}{\sin{x}+1}$

$\lim_{x \to \frac{\pi}{2}} \frac{\sin^2{x}-1}{\cos{x}(\sin{x}+1)}$

$\lim_{x \to \frac{\pi}{2}} \frac{-\cos^2{x}}{\cos{x}(\sin{x}+1)}$

$\lim_{x \to \frac{\pi}{2}} \frac{-\cos{x}}{\sin{x}+1} = \frac{0}{2} = 0$