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.