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The assignment I got is to solve the limit below using l'hospital's rule.

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

What I did was turn it into a quotient

$$ \lim_{x\to\infty} \frac{e^{x}}{e^{x^2}} $$

So now we have the indeterminate form $\frac{\infty}{\infty}$ and apply l'hospital's rule

$$ \lim_{x\to\infty} \frac{e^{x}}{2 e^{x^2}x} $$

I re-applied it a few times, but it appears that it cannot be solved this way.

My question is how can this be solved with l'hospital's rule? Please provide explanation or hint on how it can be worked out, I don't need a plain answer.

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  • $\begingroup$ $x^2$ grows faster than $x$ thus the searched limit is $0$ $\endgroup$ – Dr. Sonnhard Graubner Sep 21 '16 at 18:04
  • $\begingroup$ the circumstance is that I have to apply this rule, even if there are easier ways of finding this limit $\endgroup$ – ttdijkstra Sep 21 '16 at 18:10
  • $\begingroup$ While Dr Sonnhard avoids the instruction that L'Hospitals Rule is to be used, I agree with him that this limit can be done much easier, as he indicated. This problem is not really suitable for Hospital, nor does it show the effectiveness of an otherwise beautiful rule. Why on earth does your teacher want you to do this problem with that rule? There are other (easier) ways to do this limit. I think it is legitimate to ask $\endgroup$ – imranfat Sep 21 '16 at 18:11
  • $\begingroup$ Does the assignment make additional assumptions, e.g about the existence of the limit? $\endgroup$ – aventurin Sep 21 '16 at 19:04
  • $\begingroup$ No mention about its existence. The only instruction given was "Use l’Hospital’s rule to compute the limits" and a couple of limits like above. To give some background, this is part of a first year calculus course for chemistry. $\endgroup$ – ttdijkstra Sep 21 '16 at 19:19
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While we do not use L'Hospital's Rule herein, one can proceed by simply noting that

$$e^{x-x^2}=e^{-(x-1/2)^2}e^{1/4}=\frac{e^{1/4}}{e^{(x-1/2)^2}}$$

Then, using the elementary inequality $e^x\ge 1+x$, we can write

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

whereupon application of the squeeze theorem yields the coveted limit

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

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  • $\begingroup$ Ok, now I had to post mine too:) The more methods, the better... $\endgroup$ – imranfat Sep 21 '16 at 19:53
  • $\begingroup$ Silly mistake, I deleted my post, I don't think it adds to the OP's problem $\endgroup$ – imranfat Sep 21 '16 at 20:44
  • $\begingroup$ @imranfat No worry; I will delete this comments to avoid possible confusion of future readers. $\endgroup$ – Mark Viola Sep 21 '16 at 22:16
  • $\begingroup$ Interestingly, the same problem is asked again under a different name, but I suspect by the same person. This time no Hospital Rule: math.stackexchange.com/questions/1936212/… $\endgroup$ – imranfat Sep 21 '16 at 23:35
  • $\begingroup$ @imranfat we probably attend the same university class and have the same homework assignment $\endgroup$ – ttdijkstra Sep 22 '16 at 8:15
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As mentioned in the comments, there are much easier ways of finding the limit. However, if you have to use this calculation, let $a = \lim_{x \to \infty} e^{x-x^2}$. Your calculation gives you an algebraic equation you can use to determine exactly what the value of $a$ is.

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  • $\begingroup$ I don't see how I can proceed to solve it using this information $\endgroup$ – ttdijkstra Sep 21 '16 at 18:16
  • $\begingroup$ You arrive at the expression $a=\frac{a}{2x}$ so what happens if $x$ goes to infinity? $\endgroup$ – imranfat Sep 21 '16 at 18:17
  • $\begingroup$ You need to show that $a$ exists first before proceeding down this path. ;-)) $\endgroup$ – Mark Viola Sep 21 '16 at 18:18
  • $\begingroup$ Yeah - but we can cheat a little in a first year calculus course :P $\endgroup$ – Jacob Denson Sep 21 '16 at 18:19
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    $\begingroup$ I agree with Dr MV. This approach is not clean in spite of yielding the answer. And I wonder why the teacher wants the students to do this exercise with Hospital's rule. $\endgroup$ – imranfat Sep 21 '16 at 18:22

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