How can you solve the limit without using L'Hopital's Rule?

[Jameson]

Evaluate the following limit without using L'Hopital's Rule.

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

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) Chris L T521

Partial credit goes to kaliprasad.

Solution (from anemone):

Spoiler

Note that every exponential function is a power of the natural exponential function, or $a^x=e^{x\ln a}$, thus we can rewrite the problem as

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

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim\limits_{x \to 0} \frac{e^{x\ln 9}-e^0+e^0-e^{x\ln 5}}{x}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim\limits_{x \to 0} \left( \frac {e^{x\ln 9}-e^0}{x-0}\right)-\lim\limits_{x \to 0}\left(\frac{e^{x\ln 5}-e^0}{x-0}\right)$$ but $$\left(\frac{dy}{dx}|_{x=0}=f(0)=\lim\limits_{x \to 0}\frac{f(x)-f(0)}{x-0}\right)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{d}{dx}(e^{x\ln 9})|_{x=0}- \frac{d}{dx}(e^{x\ln 5})|_{x=0}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln 9(e^{x\ln 9})|_{x=0}- \ln 5(e^{x\ln 5})|_{x=0}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln 9(e^{0\ln 9})- \ln 5(e^{0\ln 5})$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln 9(e^0)- \ln 5(e^0)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln 9(1)- \ln 5(1)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln \frac{9}{5}$$

Alternate solution (from MarkFL):

Spoiler

First, let's rewrite the limit as follows:

$$L=\lim_{x\to0}\left(\frac{e^{x\ln(9)}}{x}-\frac{e^{x\ln(5)}}{x} \right)$$

Next, we may employ the Maclaurin series:

$$e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$

and so we find:

$$\frac{e^{ax}}{x}= \frac{1}{x}\sum_{k=0}^{\infty} \frac{(ax)^k}{k!}=\frac{1}{x}+a+ax \sum_{k=2}^{\infty} \frac{(ax)^{k-1}}{k!}$$

Applying this to the limit, we may write:

$$L=\lim_{x\to0}\left(\left(\frac{1}{x}+\ln(9)+\ln(9)x \sum_{k=2}^{\infty} \frac{(\ln(9)x)^{k-1}}{k!} \right)-\left(\frac{1}{x}+\ln(5)+\ln(5)x \sum_{k=2}^{\infty} \frac{(\ln(5)x)^{k-1}}{k!} \right) \right)$$

Simplify:

$$L=\lim_{x\to0}\left(\ln(9)+\ln(9)x \sum_{k=2}^{\infty} \frac{(\ln(9)x)^{k-1}}{k!}-\ln(5)-\ln(5)x \sum_{k=2}^{\infty} \frac{(\ln(5)x)^{k-1}}{k!} \right)$$

$$L=\lim_{x\to0}\left(\ln\left(\frac{9}{5} \right)+x\left(\ln(9) \sum_{k=2}^{\infty} \frac{(\ln(9)x)^{k-1}}{k!}-\ln(5) \sum_{k=2}^{\infty} \frac{(\ln(5)x)^{k-1}}{k!} \right) \right)$$

Hence, we find:

$$L=\ln\left(\frac{9}{5} \right)$$

Alternate solution 2 (from Chris L T521):

Spoiler

I will probably use a not-so-conventional technique, but the crux of my argument relies on using Newton's generalized binomial theorem, which says that if $x$ and $y$ are any real numbers with $|x|>|y|$ and $r$ is any complex number, then
\[(x+y)^r = \sum_{k=0}^{\infty}\binom{r}{k} x^{r-k}y^k\]
where
\[\binom{r}{k} = \frac{r(r-1)(r-2)\cdots (r-k+1)}{k!}\]
denotes the falling factorial with $\displaystyle\binom{r}{0}=1$. This generalized binomial theorem will be applied to the term $9^x$ since $9^x=(5+4)^x$. In this state, we satisfy all the necessary conditions to see that
\[\begin{aligned} 9^x &=(5+4)^x\\ & = \sum_{k=0}^{\infty}\binom{x}{k}5^{x-k}4^k \\ &= 5^x \sum_{k=0}^{\infty}\binom{x}{k} \left(\frac{4}{5}\right)^k\end{aligned}\]
In addition to all of this, my solution will also depend on knowing the Taylor series for $\ln(1+x)$:
\[\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{k}.\]
Knowing these two things, we see that
\[\begin{aligned} \lim_{x\to 0} \frac{9^x-5^x}{x} &= \lim_{x\to 0}\frac{\displaystyle\left(5^x \sum_{k=0}^{\infty}\binom{x}{k} \left(\frac{4}{5}\right)^k \right)-5^x}{x} \\ &= \lim_{x\to 0} \frac{5^x}{x}\left[\left( \sum_{k=0}^{\infty} \binom{x}{k} \left(\frac{4}{5}\right)^k\right) - 1 \right]\\ &= \lim_{x\to 0}\frac{5^x}{x} \left[\left( 1 + \frac{4}{5}x + \left(\frac{4}{5}\right)^2\frac{x(x-1)}{2!} + \left(\frac{4}{5}\right)^3\frac{x(x-1)(x-2)}{3!}+\cdots \right) - 1\right] \\ &= \lim_{x\to 0}\frac{5^x}{x} \left[\frac{4}{5}x + \left(\frac{4}{5}\right)^2\frac{x(x-1)}{2!} + \left(\frac{4}{5}\right)^3\frac{x(x-1)(x-2)}{3!}+\left(\frac{4}{5}\right)^4 \frac{x(x-1)(x-2)(x-3)}{4!} +\cdots\right]\\ &= \lim_{x\to 0}\frac{5^x}{x} \left[x\left(\frac{4}{5} + \left(\frac{4}{5}\right)^2\frac{(x-1)}{2!} + \left(\frac{4}{5}\right)^3\frac{(x-1)(x-2)}{3!}+\left(\frac{4}{5}\right)^4 \frac{(x-1)(x-2)(x-3)}{4!} +\cdots\right) \right] \\ &= \lim_{x\to 0} 5^x \left[\frac{4}{5} + \left(\frac{4}{5}\right)^2\frac{(x-1)}{2!} + \left(\frac{4}{5}\right)^3\frac{(x-1)(x-2)}{3!}+\left(\frac{4}{5}\right)^4 \frac{(x-1)(x-2)(x-3)}{4!} +\cdots \right]\\ &= \frac{4}{5} + \left(\frac{4}{5}\right)^2\frac{(-1)}{2!} + \left(\frac{4}{5}\right)^3\frac{(-1)(-2)}{3!}+\left(\frac{4}{5}\right)^4 \frac{(-1)(-2)(-3)}{4!} +\cdots \\ &= \frac{4}{5} + \frac{(-1)^1}{2} \left(\frac{4}{5}\right)^2 + \frac{2!(-1)^2}{3!} \left(\frac{4}{5}\right)^3 + \frac{3!(-1)^3}{4!} \left(\frac{4}{5}\right)^4+\cdots \\ &= \frac{(-1)^{1-1}}{1}\left(\frac{4}{5}\right)^1 + \frac{(-1)^{2-1}}{2} \left(\frac{4}{5}\right)^2 + \frac{(-1)^{3-1}}{3} \left(\frac{4}{5}\right)^3 + \frac{(-1)^{4-1}}{4} \left(\frac{4}{5}\right)^4 +\cdots \\ &= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \left(\frac{4}{5}\right)^k \\ &= \ln\left(1+\frac{4}{5}\right) \\ &= \ln\left(\frac{9}{5}\right)\end{aligned}\]
Thus, we have that
\[\lim_{x\to 0}\frac{9^x-5^x}{x}=\ln\left(\frac{9}{5}\right).\]