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I know that $\displaystyle\lim_{x\to +\infty} e^x-x=+\infty$,

$e^x$ reaches infinity much faster than $x$, and it is clear from the graph of the function that this is the case.

But I was wondering, why? What is the rule to calculate this limit, I mean because $\displaystyle\lim_{x\to +\infty} e^x-x=\infty-\infty$ which is an indeterminate form.

Fareed Abi Farraj
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5 Answers5

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Note that $e^x-x=e^x\left(1-\frac x{e^x}\right)$. So, and since $\lim_{x\to\infty}e^x=\infty$ and $\lim_{x\to\infty}\frac x{e^x}=0$, $\lim_{x\to\infty}e^x-x=\infty$.

José Carlos Santos
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You have (at least) two ways of going about this.

You could try to "factor out" a part of the function that goes to infinity, while making sure that what is left does not go to $0$. For example, you can factor out $e^x$ to get $$e^x-x = e^x(1 - xe^{-x})$$ or you could factor out $x$, and get $$e^x-x =x\left(\frac{e^x}{x} - 1\right)$$ in both cases, you get something that grows beyond bounds times something that does not converge to zero (in the second case, actually, you get something that grows beyond bounds times something that also grows beyond bonds)

Alternatively, you could use the fact that $e^x$ grows, in the limit, faster than any polynomial, and just estimate $e^x - x \geq p(x) - x$. Picking the right polynomial (i.e., anything that is at least a quadratic function) will mean that $p(x)-x$ grows beyond bounds.

Note that both methods above work not only to solve your particular problem, but a more general one, which you can try to solve for practice:

If $p$ is any polynomial, then $$\lim_{x\to\infty} e^x - p(x) = \infty.$$

Gary
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5xum
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Note that : $e^x= 1+x +{x^2}/2! +x^3/3!+...x^k/k!+...$
We have $e^x-x=1+x^2/2!+x^3/3!+...x^k/k!+...\to \infty$ as $x\to \infty$

Koro
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Product Law

Hence: $\displaystyle\lim_{x\to +\infty} e^x-x= \displaystyle\lim_{x\to +\infty} x*\dfrac{e^x-x}{x} = \displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}\dfrac{e^x}{x}-1 $

Apply L'Hopital on the second limit like you requested to get

$\displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}\dfrac{e^x}{x}-1 = \displaystyle\lim_{x\to +\infty}x* \displaystyle\lim_{x\to +\infty}{e^x}-1 $

This is now in the form $\infty*\infty$ and obviously $=\infty$

Sid
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  • I think this answer can be fixed however, you link limit product law for two existing limits . You cannot use it just like that without any comment when limit of $x$ simply do not exist. – Marek Kryspin Jun 19 '22 at 16:51
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Instead of studying $e^x-x$ we can study $\dfrac{e^x}{x}$, indeed if $x=o(e^x)$ then $e^x-x\sim e^x\to\infty$.

Let $f(x)=\dfrac{e^x}x$ then $f'(x)=\dfrac{e^x(x-1)}{x^2}\ge 0$ for $x\ge 1$

Since $f\nearrow$ then $f(x)\ge f(1)=e,\ \forall x\ge 1$.

Now using the additive to multiplicative exponential formula we get (for $x\ge 2$):

$$f(x)=\dfrac{e^x}x=\dfrac{e^{\frac x2}e^{\frac x2}}{x}=\frac 12f(\tfrac x2)\cdot e^{\frac x2}\ge \frac e2e^{\frac x2}\to+\infty$$

zwim
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  • What does the arrow pointing to the top right mean? Does it mean that $f$ is increasing? – A-Level Student Mar 01 '21 at 14:07
  • yes exactly, increasing \nearrow and decreasing is \searrow. – zwim Mar 01 '21 at 14:51
  • I see, thanks. Is that standard notation? – A-Level Student Mar 01 '21 at 15:38
  • Well I don't know, but tables of variation like this one https://calque.pagesperso-orange.fr/latex/captlatex/latextab.png are very common in studying functions, so the idea of using the arrows as shortcuts (in plain text outside of the table) is quite natural I guess. – zwim Mar 01 '21 at 16:16