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Topic: Another Limit Problem
Replies: 4   Last Post: Feb 11, 2013 11:48 AM

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Charles Hottel

Posts: 10
Registered: 9/5/11
Re: Another Limit Problem
Posted: Jan 29, 2013 12:06 PM
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"A N Niel" <> wrote in message
> In article <>, Charles
> Hottel <> 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.

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