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I have never come across this type of problem so far and am lost on how to proceed. I was told never to plug in infinity because it can get theoretical. $\infty-\infty=0$? $e^0=0$?

Problem: $$\lim_{x\to \infty}e^{x-x^2}$$

projectilemotion
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Itsreason
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  • Note: $\infty-\infty$ isn't zero, it can be any number you'd want it to be, so it is not a useful concept. Also $e^0=1$, not $0$. – John Doe Apr 25 '17 at 18:22
  • What happens to the expression at x=1? What happens at x=2? At x=3? Keep increasing x? Is the value of the expression monotonically decreasing? – user2808118 Apr 25 '17 at 18:24
  • Note that $e^{x-x^2}=e^{x^2(\frac{1}{x}-1)}$ – MasB Apr 25 '17 at 18:25

5 Answers5

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Note that $\lim_{x \to \infty} x - x^2 = -\infty$ so if we set $u = x - x^2$ we have

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

levap
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  • As a matter of courtesy, since your answer is more complete than mine, should I delete my answer? – Michael Stachowsky Apr 25 '17 at 18:25
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    @MichaelStachowsky: Nah, it's OK. It often happens that people post answers at the same time that looks similar. Let the OP decide what he prefers. – levap Apr 25 '17 at 18:29
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When evaluating a limit, you do not, strictly speaking, "plug in infinity" (although you'll be using your intuition to do so). You are correct. Consider the following:

$$lim_{x \rightarrow \infty}(x-x^2)$$

In this case, we see that the limit is $-\infty$. Then, $e^{-\infty}$ = 0, as you specified.

Michael Stachowsky
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${lim_{x \rightarrow +\infty}x-x^2}=lim_{x \rightarrow +\infty}(-x^2)=-\infty$
$lim_{x \rightarrow +\infty}e^{x-x^2}=e^{-\infty}=0$

cxaxm
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$ \lim_{x\to \infty}e^{x-x^2}=\lim_{x \to \infty}\frac{e^x}{e^{x^2}}=\lim_{x \to \infty}\frac{1}{e^{x^2-x}}=\lim_{x \to \infty}\frac{1}{e^{x(x-1)}}=\frac{1}{\infty}=0$

kmitov
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Using the inequality $e^x\ge 1+x$, which I showed in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality, we can write

$$0\le e^{x-x^2}=\frac{1}{e^{x^2-x}}\le \frac{1}{1+x^2-x}$$

whereupon applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}e^{x-x^2}=0}$$

Mark Viola
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