# Thread: Calculus Limit Problem

1. ## Calculus Limit Problem

I have a Calculus project. I have to find the following limit without using L'Hospital:

Limit as x approaches a (x->a) of (cos(x)-cos(a))/(x-a)

2. Originally Posted by djo201
Thanks for the fast reply but I'm not getting it. Can you explain a little bit in more detail?

Thanks

Edit: I know the derivative of cos(x) or cos(a) is -sin(x) or -sin(a), respectively, but how do I get to that result?
by defintion

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

here you have $f(x) = \cos x$

so you can just say it is the derivative of cos(a) and just state that as the limit, namely, -sin(a)

if you would rather calculate it manually, note that an alternate definition for the derivative is as follows

$f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

it is easier to find this particular limit in that form

getting your limit in that form is not as hard as you think

let $x = a + h$, then $\lim_{x \to a} \frac {\cos x - \cos a}{x - a}$ becomes $\lim_{h \to 0} \frac { \cos (a + h) - \cos a}h$

now proceed using the addition formula for cosine to simplify

(note, you will need your special trig limits here)

3. Originally Posted by Jhevon
by defintion

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

here you have $f(x) = \cos x$

so you can just say it is the derivative of cos(a) and just state that as the limit, namely, -sin(a)

if you would rather calculate it manually, note that an alternate definition for the derivative is as follows

$f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

it is easier to find this particular limit in that form

getting your limit in that form is not as hard as you think

let $x = a + h$, then $\lim_{x \to a} \frac {\cos x - \cos a}{x - a}$ becomes $\lim_{h \to 0} \frac { \cos (a + h) - \cos a}h$

now proceed using the addition formula for cosine to simplify

(note, you will need your special trig limits here)
Thank you very much. Now I get it.

4. Ok, one last question, why does x->a become h->0?

5. Originally Posted by djo201
Ok, one last question, why does x->a become h->0?
$x = a + h$

$\Rightarrow h = x - a$

now as $x$ gets close to $a$, what does $h$ get close to?

6. Originally Posted by Jhevon
$x = a + h$

$\Rightarrow h = x - a$

now as $x$ gets close to $a$, what does $h$ get close to?
Thanks again.