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The limit below is solved using multiple methods. $$\lim_{x\to\infty} \frac{x^n}{e^x}$$

However, I am trying to solve it using the comment made below the question:

Only one application of l'Hopital's rule is necessary if you take logarithms first.

$$\lim_{x\to\infty} \frac{x^n}{e^x} =\lim_{x\to\infty}\exp \left(\ln \frac{x^n}{e^x}\right) =\lim_{x\to\infty}\exp \left(n\ln x-x\right)$$

And now I don't know how to proceed since this is the $\infty-\infty$ case. What am I missing, or was something else meant by the comment?

Starlight
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  • $\ln x$ grows much slower than $x$, so e.g $ n\ln x -x< -x/2$ for large $x$. (To prove it, you may consider the limit of $x/\ln x$). – van der Wolf Feb 03 '25 at 06:33
  • @vanderWolf I knew that ln $x$ grows much slower than x, but did not know how to proceed from there. Even now, how do you formally arrive at the inequality you have got. – Starlight Feb 03 '25 at 06:45
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    By L'Hopital rule $\lim_{x\to\infty} \frac{x}{\ln{x}}=\lim_{x\to\infty} \frac{1}{1/x}=\lim_{x\to\infty} x=+\infty$ which imples that $2n\ln x< x$ for all large enough $x$. As a result, $n\ln x -x\to -\infty$, and $\exp(n\ln x -x)\to 0$ as $e^{-\infty}=0$. – van der Wolf Feb 03 '25 at 10:02
  • @Starlight I have prepared a second answer that combines L'Hôpital's rule, which is applied only once (as you intended), with Mathematical induction to evaluate the original limit . –  Feb 05 '25 at 19:23
  • @Starlight Was it something you didn't like about my answer which included induction using L'Hôpital only once, which is not usually common for limits ? –  Feb 19 '25 at 04:23
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    @brokenspirit I actually didn't pay too much attention to the argument since it was only applicable to integer values. But I think as an answer, it's great. – Starlight Feb 19 '25 at 07:00

4 Answers4

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Generally, a $\infty-\infty$ indeterminate form - that is, we're investigating $$ \lim_{x \to c}\left(f(x) - g(x)\right) \text{ where } \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = \infty $$ can be transformed into something for which L'Hopital's rule is applicable by the following rules: $$ \lim_{x \to c}\left(f(x) - g(x)\right) = \lim_{x \to c} \frac{\frac{1}{g(x)} - \frac{1}{f(x)}}{\frac{1}{f(x)g(x)}} = \lim_{x \to c} \ln \left( \frac{e^{f(x)}}{e^{g(x)}} \right) $$ where the former is a $0/0$ case and the latter is an $\infty/\infty$ case.

Taking the first as our inspiration, then, $$ \lim_{x \to \infty}\left(n \ln x - x\right) = \lim_{x \to \infty} \frac{ \frac{1}{x} - \frac{1}{n \ln x}}{\frac{1}{n x \ln x}} = \lim_{x \to \infty} \frac{ \frac{n}{x} - \frac{1}{ \ln x}}{\frac{1}{ x \ln x}} $$ Apply L'Hopital's: $$ \lim_{x \to \infty} \frac{ \frac{n}{x} - \frac{1}{ \ln x}}{\frac{1}{ x \ln x}} = \lim_{x \to \infty} \frac{ -\frac{n}{x^2} - \frac{1}{ x \ln^2 x}}{-\frac{1+\ln x}{ x^2 \ln^2 x}} = \lim_{x \to \infty} \left( n \ln x + \frac{n+x}{1+\ln x} - n \right) $$ from which the result is obvious.

... personally I'd sooner take a bunch of derivatives of $x^n$ and then "notice a pattern," versus ever deal with the quotient rule. One I can do in my head, the other I can't lol.

PrincessEev
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    Where did you learn this? I always attack these kinda limits by doing $x\to\frac{1}{x}$ – Blue Cat Blues Feb 03 '25 at 14:55
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    Derived it once and threw the expressions in my notes. Nothing elegant, but simple to retrieve when I recognize a suitable application for which simpler methods probably aren't allowed. – PrincessEev Feb 03 '25 at 16:52
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Technically, do you want to solve the limit by applying L'Hopital's rule only once ?

If so, not only for natural numbers $n$, but in general for all positive $n>0$, you can write that :

$$\begin{align}\lim_{x\to\infty}\frac{x^n}{e^x}&=\lim_{x\to\infty}\left(\frac{x}{e^{\frac {x}{n}}}\right)^n\\ &=\left(\lim_{x\to\infty}\frac{x}{e^\frac{x}{n}}\right)^n \color {blue} {\text{Interchanging limit and power}}\\ &=\left(\lim_{x\to\infty}\frac{1}{\frac {1}{n}\cdot e^{\frac{x}{n}}}\right)^n \color {blue} {\text{Applying L'Hopital}} \\ &=0^n=0\,.\end{align}$$

Starlight
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  • That's a great solution. – Starlight Feb 03 '25 at 12:42
  • @Starlight Thank you. Here, we used the fact below :

    $${\lim_{x\to\infty}f(x)=L\in\Bbb R_{\ge 0}\implies\lim_{x\to\infty}\left(f(x)\right)^n=L^n}$$

     –  Feb 03 '25 at 16:52
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True to the author's intention, we can also make a solution using mathematical induction by applying L'Hopital's rule only once .

For $n=0$, trivially we are done .

Suppose that ,

$$\lim_{x\to\infty}\frac {x^k}{e^x}=0$$

is true for some integer $n=k>0\,.$

So, we have :

$$\begin{align}\lim_{x\to\infty}\frac{x^{k+1}}{e^x}&=\lim_{x\to\infty}\frac{x}{\frac {e^x}{x^{k}}}\\ &=\lim_{x\to\infty}\frac {1}{e^x x^{-k} \left(1 - k/x\right)}\\ &=\lim_{x\to\infty}\frac{x^k}{e^x\left(1-k/x\right)}\\ &=\lim_{x\to\infty}\frac{x^k}{e^x}\cdot\lim_{x\to\infty}\frac{1}{1-k/x}\\ &=0\cdot 1=0\;.\end{align}$$

By mathematical induction, we are done .

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If you take $u(x) = n\ln x -1$ and $v(x) = \frac 1 x$, then $u'(x)=\frac n x$ and $v'(x)=-\frac 1 {x^2}$, so $\frac {u'(x)}{v'(x)} = -nx$, and the limit of $e^{-nx}$ is $0$.

Acccumulation
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  • How does this relate to the question at hand? We have $x^n$ over $e^x$. In particular how are $u(x)$ and $v(x)$ related to the question. – Starlight Feb 04 '25 at 02:12