Date: Jan 29, 2013 12:06 PM
Author: Charles Hottel
Subject: Re: Another Limit Problem

"A N Niel" <anniel@nym.alias.net.invalid> wrote in message

news:290120130712519075%anniel@nym.alias.net.invalid...

> In article <LbydnWRm6eYGV5rMnZ2dnUVZ_uKdnZ2d@earthlink.com>, Charles

> Hottel <chottel@earthlink.net> wrote:

>

>> How can I find lim(x->0) tan 5x /sin 2 x ?

>>

>> I have tried manipulating this every way I can think of, but I still end

>> up

>> with a fraction that contains a denominator of zero. Thanks.

>>

>>

>

> Can you do lim(x->0) (sin x)/x ?

> If so, think of how that may be relevant to your problem.

I know that lim(x->0) (sinx/x) = 1 and I was just about to reply that it

does not help when I actually figured out a way that it does. Problem

solved .Thanks.