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



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



