"A N Niel" <firstname.lastname@example.org> wrote in message news:email@example.com... > In article <LbydnWRm6eYGV5rMnZ2dnUVZ_uKdnZ2d@earthlink.com>, Charles > Hottel <firstname.lastname@example.org> 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.