lim (2^x-3^x)/(cos(x)-1)

Question

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?

Answer 1

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)