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.
Best Answer
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.