lim (2^x-3^x)/(cos(x)-1)
x->0
The answer the book has for this is "diverges." I know I'm supposed to use L'Hospitals' rule, but I'm not sure how. Can you walk me through the steps?
it's quite simple. The rule states that
lim f(x)/g(x) = lim f'(x)/g'(x)
so, your limit is the same as
(ln2*2^x-ln3*3^x)/-sinx(x)
now, as x->0, that -> (ln2 - ln3)/0 = ∞
This is confirmed by the graph at
http://www.wolframalpha.com/input/?i=(2%5Ex-3%5Ex)%2F(cos(x)-1)