A Tricky Limit: $\lim \limits_{x\rightarrow 0} \frac{9^x-5^x}{x}$ without L'Hospital 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. 
 A: 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 $\exp (x)$.
A: 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.
A: 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.
A: You might also consider
$$\lim\limits_{x \to 0} \frac{9^x-5^x}{x}$$
$$\lim\limits_{x \to 0} \frac{9^x-1}{x} -\frac{5^x-1}{x}$$
$$\lim\limits_{x \to 0} \frac{9^x-9^0}{x-0} -\frac{5^x-5^0}{x-0}=\log \frac{9}{5}$$
A: Remember that $\lim_{y\to 0}\frac{e^y-1}{y}=1$. Set $y=\log(9/5)\cdot x \;\;$ and note that
\begin{align}
\lim_{x\rightarrow 0} \frac{9^x-5^x}{x}
=
&
\lim_{x\rightarrow 0}\;\; 
(5^x)
\cdot
\frac{(9/5)^x-1}{x}
\\
=
&
\lim_{x\rightarrow 0}\;\; 
(5^x)
\cdot
\frac{e^{x\log(9/5)}-1}{x}
\\
=
&
\lim_{x\rightarrow 0}\;\; 
(\log(9/5)\cdot 5^x)
\cdot
\frac{e^{x\log(9/5)}-1}{x\cdot \log(9/5)}
\\
=
&
\lim_{x\rightarrow 0}\;\; 
(\log(9/5)\cdot 5^x)
\lim_{x\rightarrow 0}
\frac{e^{x\log(9/5)}-1}{x\cdot \log(9/5)}
\\
=
&
\log(9/5)\cdot\lim_{x\rightarrow 0}\;\; 
(5^x)
\lim_{y\rightarrow 0}
\frac{e^{y}-1}{y}
\\
=
&
\log(9/5)\cdot 5^{0}\cdot \lim_{y\rightarrow 0}
\frac{e^{y}-1}{y}
\\
=
&
\log(9/5)
\\
\end{align}
A: In addition to the other good answers, you can also try this (or those who might be interested):


*

*Since $a^x \to 1+x \ln{a}$, when $x\to 0$.

*Your limit might be rewritten as follows:
$$L = \lim_{x\to0} \frac{1+x \ln 9 - 1 - x \ln 5}{x} = \lim_{x \to 0} \frac{x \ln{(9/5)}}{x} = \ln(9/5).$$


Cheers.
