How to evaluate $\lim\limits_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}$ without L'Hôpital's rule? 
Evaluate: $$\lim\limits_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}.$$

This is an exercise problem from my book. I couldn't solve the problem without using L'Hôpital's rule. Here is my solution that uses L'Hôpital's rule:
Since we achieve $\frac00$ in the limit, we can use L'Hôpital's rule. Then, we have
$\lim\limits_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}$ $=\lim\limits_{x\to 0}\frac{5e^{5x}-9(1+x)^8}{\frac{1}{1+x}}\\ =(1+0)(5e^0-9\cdot 1^8)\\ =-4$
I need a solution that doesn't use L'Hôpital's rule.
 A: \begin{align}
\lim_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}&=\lim_{x\to 0}\frac{x}{\ln(1+x)-\ln1}\left(\frac{e^{5x}-1}{x}-\frac{(1+x)^9-1}{x}\right)\\
&=\frac1{\left.\left(\ln(1+x)\right)'\right|_{x=0}}\left(\left.\left(e^{5x}\right)'\right|_{x=0}-\left.\left((1+x)^9\right)'\right|_{x=0}\right)\\
&=5-9\\&=-4
\end{align}
A: Just use Taylor Expansion, $$e^{5x}=1+5x+\frac{25}{2!}x^2+\frac{125}{3!}x^3+....$$
$$(1+x)^9=1+9x+36x^2+^9C_3x^3+^9C_4x^4+....$$
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$
Plugging in the above equation $1$ cancels out in the numerator and taking  $x$ common from numerator and denominator and cancelling it out gives answer=$-4$ as rest of the terms have a power of $x$ and become =$0$ on applying the limit.
Note: I didn't write the further terms to save time as anyways all powers of $x$ will vanish on applying the limits.
A: I'll assume that you can use some "known" limits...
\begin{align*}
\lim_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}=&\lim_{x\to 0}\dfrac{e^{5x}-(1+9x+36x^2+ \cdots)}{\ln(1+x)}=\lim_{x\to 0}\dfrac{e^{5x}-1}{5x}\dfrac{5x}{\ln(1+x)} -\\
& \lim_{x\to 0}\dfrac{9x}{\ln(1+x)} -\lim_{x\to 0}\frac{36 x^2 + \cdots}{\ln(1+x)} = 5 -9 -0 = -4
\end{align*}

$$
\lim_{x\to 0} \frac{e^x-1}{x} = 1
$$
$$
\lim_{x\to 0}\frac{\ln(1+x)}{x} = 1 
$$
A: Another way:
$$\begin{aligned}\lim_{x\to 0}\frac{e^{5x}-(1+x)^9}{\ln(1+x)}&=\lim_{x\to 0}\frac{e^{5x}-1+1-(1+x)^9}{\ln(1+x)}\\&=\lim_{x\to 0}\left(\frac{e^{5x}-1}{5x}\cdot\frac{5x}{\ln(1+x)}-\frac{(1+x)^9-1}{\ln(1+x)}\right)\\&=\lim_{x\to 0}\left(\frac{e^{5x}-1}{5x}\frac{5x}{\ln(1+x)}-\frac{e^{9\ln(1+x)}-1}{9\ln(1+x)}9\right)\\&=5-9\\&=-4\end{aligned}.$$
I would also recommend taking a look at this somewhat related answer.
