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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.

Oshawott
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4 Answers4

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\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}

Kay K.
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    You are cheating a little bit by using a "personalized" proof of L'Hôpitals rule instead of the rule itself. – PierreCarre Jul 20 '21 at 16:44
  • This is a clever answer but I can't imagine this is what the question asker (professor, textbook author, not OP) had in mind – cohomonoid Jul 20 '21 at 16:45
  • @PierreCarre This usually doesn't work in general, because when the numerator is separated into sum of two terms then usually each of two terms doesn't converge individually. But, in this case it does. So there's no reason to not do this way. – Kay K. Jul 20 '21 at 16:49
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    @KayK. Don't get me wrong, I like this approach... – PierreCarre Jul 20 '21 at 16:56
  • But you don't need to separate the terms right? Just write $e^{5x} - (1+x)^9 = x \left( \frac{e^{5x} - (1+x)^9 - 0}{x}\right)$. – Arctic Char Jul 20 '21 at 17:00
  • @ArcticChar Yeah I just realized after I saw PierreCarre's solution. – Kay K. Jul 20 '21 at 17:01
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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.

UNAN
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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*}

$$ \lim_{x\to 0} \frac{e^x-1}{x} = 1 $$ $$ \lim_{x\to 0}\frac{\ln(1+x)}{x} = 1 $$

PierreCarre
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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.

PinkyWay
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