# special limits of trig functions

• Nov 8th 2006, 12:59 PM
myoplex11
special limits of trig functions
compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)
• Nov 8th 2006, 01:42 PM
topsquark
Quote:

Originally Posted by myoplex11
compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)

First of all, it's "pi" not "pie."

I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

$\lim_{x \to \pi} \frac{cosx + 1}{x - \pi}$

Note that this is of the form 0/0, so use L'Hopital's rule:

$\lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0$

-Dan
• Nov 8th 2006, 01:52 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
First of all, it's "pi" not "pie."

I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

$\lim_{x \to \pi} \frac{cosx + 1}{x - \pi}$

Note that this is of the form 0/0, so use L'Hopital's rule:

$\lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0$

-Dan

You can do this,
$\lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}$
Using the identity,
$\cos x=-\cos (x-\pi)$
Now this is a composition function,
$f(x)=\frac{1-\cos x}{x}$
$g(x)=x-\pi$
And,
$\lim_{x\to \pi}g(x)=0$
Thus the limit of,
$\lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0$
By composition rule for limits.
That is a standard limit.
• Nov 8th 2006, 01:57 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
You can do this,
$\lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}$
Using the identity,
$\cos x=-\cos (x-\pi)$
Now this is a composition function,
$f(x)=\frac{1-\cos x}{x}$
$g(x)=x-\pi$
And,
$\lim_{x\to \pi}g(x)=0$
Thus the limit of,
$\lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0$
By composition rule for limits.
That is a standard limit.

Clever! I like it. :)

-Dan