$$\lim_{x\to\infty}e^{x-x^2} = 0$$
I'm a bit lost on this one. Graphically, it tends to 0, hence the limit is 0. But, how could I prove it algebraically?
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Welcome to Math Stack Exchange. $\lim_{x\to\infty} e^x-x^2=\infty$ – J. W. Tanner Apr 29 '19 at 01:09
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Make sure that you have the correct function. – Mohammad Riazi-Kermani Apr 29 '19 at 01:11
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@J.W.Tanner I corrected the equation. – binibiningtinamoran Apr 29 '19 at 01:11
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@MohammadRiazi-Kermani Thank you for pointing that out. I missed a couple of curly braces around the exponent. – binibiningtinamoran Apr 29 '19 at 01:12
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Why would it be $1$ in the first place, exactly? Because substitution (naively!) gives "$\infty - \infty$" in the exponent? – PrincessEev Apr 29 '19 at 01:12
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$x-x^2\to -\infty$ thus $e^{x-x^2}\to 0$ – Alexander Gruber Apr 29 '19 at 01:13
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Are you familiar with L'hopital's rule? – J. W. Tanner Apr 29 '19 at 01:13
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@EeveeTrainer This is why I am asking here. I thought $$\infty-\infty$$ = 0 – binibiningtinamoran Apr 29 '19 at 01:13
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$\infty-\infty$ is indeterminate; cf. this question – J. W. Tanner Apr 29 '19 at 01:14
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Well, intuitively, the reason why is because $x^2$ "grows faster" than $x$. So $x-x^2$ goes towards $-\infty$ instead, since $x^2$ dominates the value of the function. – PrincessEev Apr 29 '19 at 01:14
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@J.W.Tanner I am very familiar with the L'hopital's rule. However, I am doing a review on this textbook and L'hopital's rule isn't mentioned yet. – binibiningtinamoran Apr 29 '19 at 01:15
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2@binibiningtinamoran $\infty - \infty$ does not equal $0$. It is an indeterminate form, just like $\frac{\infty}{\infty}$ or $\frac{0}{0}$. – Brian61354270 Apr 29 '19 at 01:16
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$x-x^2$ isn't less than $0$ for all $x$, just for $x$ such that $x<0,x>1$, one or the other. (Of course, if $x\to\infty$, that condition is met.) Also, it's not $e$ raised to the power of anything negative, but rather $e$ to an exponent such that said exponent goes to $-\infty$ in the limit. – PrincessEev Apr 29 '19 at 01:31
3 Answers
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Intuitively, $e^{-x^2}$ goes to zero very fast. If you know $\lim_{x \to \infty}e^{-x}=0$ you can just say that $x-x^2 \lt -x$ for $x \gt 2$, so $e^{x-x^2} \lt e^{-x} \to 0$.
Ross Millikan
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$e^{x-x^2}=e^{x(1-x)}$. This means that when $x$ goes to $\infty$, the exponent goes to $\infty\cdot(-\infty)=-\infty$, and $e^{-\infty}=0$.
Andrei
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$$\lim_{x\to\infty}e^{x-x^2} = \lim_ {x\to\infty} \frac {e^x}{e^{x^2}} = \lim_ {x\to\infty} \frac {1}{e^{x^2-x}} = 1/{\infty} = 0 $$
Mohammad Riazi-Kermani
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