What is wrong in the following method of evaluating $\lim_{x \rightarrow \infty}\frac{x^{n}}{e^{x}}$ ($n$ not necessarily an integer)?

Question

Let $f(x)=\frac{x^{n}}{e^{x}}$ and $y(x)=\ln[f(x)]=\frac{n\ln x}{x}$.

Now, using L’Hospital’s rule,

$\lim_{x \rightarrow \infty}y(x)=\lim_{x \rightarrow \infty}\frac{n\ln x}{x}=\lim_{x \rightarrow \infty}\frac{n}{x}=0$.

Thus,

$\lim_{x \rightarrow \infty}f(x)=\lim_{x \rightarrow \infty}e^{y(x)}=e^{\lim_{x \rightarrow \infty}y(x)}=e^0=1$

However, the book I follow and another question on stackexchange (What is the limit $x\to\infty$ of $x^n \cdot e^{-x}$?) say that the limit is $0$. Where am I going wrong?

$\ln[f(x)]$ equals $n\ln x - x$, not $\frac{n \ln x}{x}$. – Kimmel Jul 12 '20 at 13:56 — Kimmel, Jul 12 '20 at 13:56
Oops, my bad. Thank you! – Hassan Mehmood Jul 12 '20 at 13:59 — Hassan Mehmood, Jul 12 '20 at 13:59

score 1 · Answer 1 · answered Jul 12 '20 at 13:58

1

What you did is wrong. We have $\log\bigl(f(x)\bigr)=n\log(x)-x$, and $\lim_{x\to\infty}n\log(x)-x=-\infty$. Therefore, your limit is indeed equal to $0$.

answered Jul 12 '20 at 13:58

José Carlos Santos

427,504

Yep, Kimmel in the comments above clarified. It was quite silly of me. Thanks for correcting! – Hassan Mehmood Jul 12 '20 at 14:02

What is wrong in the following method of evaluating $\lim_{x \rightarrow \infty}\frac{x^{n}}{e^{x}}$ ($n$ not necessarily an integer)?

1 Answers1

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