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

 Messages: [ Previous | Next ]
 JEMebius Posts: 913 Registered: 12/13/04
Re: Another Limit Problem-
Posted: Feb 11, 2013 11:48 AM

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.
>
>

tan(5x) / sin(2x) = [tan(5x)/x] / [sin(2x)/x] for all x <> 0;

lim(x->0) [tan(5x)/x] = lim(5x->0) [tan(5x)/x] = 5 * lim(5x->0) [tan(5x)/5x] = 5 * 1 = 5;

sin(2x)/x is treated likewise; finally lim(x->0) [tan(5x)/sin(2x)] = 5/2.

Happy studies: Johan E. Mebius

Date Subject Author
1/29/13 Charles Hottel
1/29/13 A N Niel
1/29/13 Charles Hottel
1/29/13 gnasher729
2/11/13 JEMebius