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.

  • 5
    $\begingroup$ You could recognize this as $\frac{d}{dx}\left(9^x-5^x\right)\big|_{x=0}$ (using the definition of the derivative, not L'Hospital's rule). $\endgroup$ – Nick Strehlke Sep 27 '11 at 22:56

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.

  • $\begingroup$ @Downvoter If something is not clear then please feel welcome to ask questions. $\endgroup$ – Bill Dubuque Oct 3 '15 at 23:16

You might also consider

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

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

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


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 $\exp (x)$.

  • $\begingroup$ Calculus one students wouldn't know power series, but I agree that this would work. $\endgroup$ – JSchlather Sep 27 '11 at 22:55
  • $\begingroup$ Ah, ok. I'm not from the US so I have no idea what 'Calculus 1' is :) $\endgroup$ – Chris Taylor Sep 27 '11 at 22:57
  • $\begingroup$ At my university it mostly just covers limits and derivatives with some applications and maybe integration at the end. $\endgroup$ – JSchlather Sep 27 '11 at 22:57
  • $\begingroup$ Calculus students should be aware that 1+x+x^2/2!+... differentiates to itself, and can be written as exp(x) -- it is the fundamental result. If it is lacking in a particular syllabus, it is the fault of that syllabus. $\endgroup$ – P i Jun 7 '15 at 15:39

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.


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}


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: $$L = \lim_{x\to0} \frac{1+x \ln 9 - 1 - x \ln 5}{x} = \lim_{x \to 0} \frac{x \ln{(9/5)}}{x} = \ln(9/5).$$



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