A Tricky Limit: $\lim \limits_{x\rightarrow 0} \frac{9^x-5^x}{x}$ without L'Hospital

Question

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.

Answer 1

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.

Answer 2

Use $a^x = \exp(x\ln a)$ to get

$$ \begin{align} \frac{9^x - 5^x}{x} & = \frac{\exp(x\ln 9) - \exp(x\ln 5)}{x} \end{align} $$

and then remember that power series of $\exp (x)$.

Answer 3

You could recognize this as $\frac{d}{dx}\left(9^x-5^x\right)\big|_{x=0}$. That is: $$ \begin{align*} \lim_{x\to0}\frac{9^x-5^x}{x} &= \lim_{x\to0}\left(\frac{e^{x\log{9}}-e^{0\cdot\log{9}}}{x}-\frac{e^{x\log{5}}-e^{0\cdot\log{5}}}{x}\right)\\ &= \frac{d}{dx}e^{x\log{9}}\big|_{x=0}-\frac{d}{dx}e^{x\log{5}}\big|_{x=0}, \end{align*} $$ by definition of the derivative at zero.

Answer 4

You might also consider

$$\lim\limits_{x \to 0} \frac{9^x-5^x}{x}$$

$$\lim\limits_{x \to 0} \frac{9^x-1}{x} -\frac{5^x-1}{x}$$

$$\lim\limits_{x \to 0} \frac{9^x-9^0}{x-0} -\frac{5^x-5^0}{x-0}=\log \frac{9}{5}$$