I tried to solve: $$\lim\limits_{x \to \infty} e^{x-x^2}$$ but I can't get it done. I've tried to use Hopital's rule after rewriting it to: $$\lim\limits_{x \to \infty} \frac{e^x}{e^{x^2}}$$ but this does not lead to a solition. Anyone with a good hint?

  • $\begingroup$ Perhaps you can invoke the continuity of the exponential function. $\endgroup$ – panini Sep 21 '16 at 20:31
  • $\begingroup$ You could try finding the limit of the logarithm instead. $\endgroup$ – Mosquite Sep 21 '16 at 20:39
  • $\begingroup$ Well, noting x-x^2 -> - infinity it makes sense to imagine this limit tends to 0. If I can use theorems and technics to prove it.... for large x (x > 2) I can assume -x > x - x^2 > - x^2 so I can use squeeze th. I guess. Lvhopitals rule gives me e^x/2xe^(x^2) < 1/2xx -> 0. I have options. $\endgroup$ – fleablood Sep 21 '16 at 20:41
  • $\begingroup$ Hint: Just consider $\lim\limits_{x \to \infty} x-x^2$. and apply that result to your limit. You are overcomplicating this limit big time... $\endgroup$ – imranfat Sep 21 '16 at 21:00
  • $\begingroup$ @fleablood Thanks, I indeed derived e^x/2xe^(x^2) but do not see how the step e^x/2xe^(x^2) -> 1/2xx is made (getting the e-power out of dominator and denominator). I also think 0 is correct; I used a graph to check. If you could explain that extra step, it would be fine. If you post it as an answer, I will upvote it. $\endgroup$ – Sjips Sep 21 '16 at 21:01

If $x > 2$ then $-x = x - 2x > x - x^2$ so $e^{x - x^2} < e^{-x}$ for $x > 2$. and $e^{x-x^2} > 0$ so $0 < \lim_{x\rightarrow \infty}e^{x - x^2}< \lim_{x\rightarrow \infty}e^{-x}= 0$. so $\lim_{x\rightarrow \infty}e^{x - x^2} = 0$.

... or ....

$\frac {de^x}{dx} = e^x$, $\frac {de^{x^2}}{dx} = 2xe^{x^2} > 2x e^{x}$ for $x > 1$

So by l'hopitals rule $\lim \frac {e^x}{e^{x^2}} = \lim \frac {e^x}{2xe^2} \le \lim \frac 1{2x} = 0$. So $e^{x - x^2} > 0$, $\lim e^{x- x^2} \ge 0$ so $\lim e^{x-x^2} = 0$.

.... or ....

For $\epsilon > 0$ and $x > 1 + \sqrt{\ln 1/{\epsilon}}$ then $|e^{x - x^2} - 0|= 1/e^{x^2 - x} < 1/e^{(x - 1)^2} < 1/e^{\sqrt{\ln 1/\epsilon}^2}= \epsilon$.



as $e^x$ is continuous $\lim e^{x- x^2} = e^{\lim x - x^2 = \lim -y} = 0$.

As long as you realize $x-x^2 \rightarrow -\infty$ and $\lim_{y\rightarrow -\infty} e^y = 0$ you have many options to prove it.


Note a polynomial is equivalent at infinity to its highest degree term, i.e. here: $\;x-x^2\sim_\infty -x^2$, hence $\lim_{x\to\infty}(x-x^2)=\lim_{x\to\infty}(-x^2)=-\infty$. By continuity $\;\mathrm e^{x-x^2}$ tends to $0$.

  • $\begingroup$ This argument should be made more precise. If they're equivalent at infinity, then naively one could argue $\lim \ (x - x^2) - (-x^2) = \lim \ (-x^2) - (-x^2) = 0$ which is wrong. $\endgroup$ – MathematicsStudent1122 Sep 21 '16 at 23:52
  • $\begingroup$ Every one has seen equivalence is not compatible with addition or subtraction, only with multiplication and division (and composition with logarithm, to a certain extent). I suppose students know the precise meaning of equivalence, and at least the main rules. $\endgroup$ – Bernard Sep 21 '16 at 23:57

You can mind in this way: since

$$x^2 > x$$

as $x\to \infty$, then $x - x^2 = x(1-x) \approx x\cdot(-x) = -x^2$ hence

$$e^{x -x^2} \approx e^{-x^2} = 0$$


One way forward uses only the elementary inequality

$$\bbox[5px,border:2px solid #C0A000]{e^x\ge 1+x} \tag 1$$

along with the squeeze theorem.

Proceeding, we write


Then, applying $(1)$ reveals

$$0\le \frac{e^{1/4}}{e^{(x-1/2)^2}}\le \frac{e^{1/4}}{1+(x-1/2)^2}$$

whereupon application of the squeeze theorem yields the coveted limit

$$\lim_{x\to \infty}e^{x-x^2}=0$$

  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Oct 5 '16 at 23:35

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