I'm teaching a recitation for a calculus 1 class this quarter and through some miscommunication I was under the impression that I needed to present a method to finding the limit of
$$\lim_{x\rightarrow 0} \frac{9^x-5^x}{x}$$
without using L'Hospital's rule. I found rather quickly, much to my annoyance, that I was unable to find the limit without applying L'Hospital's rule. I asked several of my friends who were also unable to solve it. I was wondering if there was an elementary solution to such a limit, that is something understandable by a beginning calculus 1 student.
Edit: To be more clear the students in my recitation have only just learned limits and haven't even reached derivatives yet.
Best Answer
Hint $\ $ Rewrite it as $\displaystyle\ \ 5^x \dfrac{(9/5)^x-1}{x}\ =\ 5^x \dfrac{{\it e}^{\:cx}-1}{x}\ $ for $\,\ c = \log(9/5).$
The limit of the latter fraction is well-known - with various proofs, e.g. by power series, or by recognizing it as a first derivative. See my prior posts for many further examples of the latter.