Asked by Lisa
Evaluate the limit using l'hospitals rule. lim->0 (3^x-6^x)/x. Do not know where to even begin on this one... please help.
Answers
Answered by
Steve
If you don't even know where to begin, you must not have read the section on this topic. It says that
the limit of f(x)/g(x) = f'(x)/g'(x)
so, your limit is just
(ln3 * 3^x - ln6 * 6^x)/1
= (ln3-ln6)
= ln(3/6) = -ln2
the limit of f(x)/g(x) = f'(x)/g'(x)
so, your limit is just
(ln3 * 3^x - ln6 * 6^x)/1
= (ln3-ln6)
= ln(3/6) = -ln2
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