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We are asked to evaluate the limit $$\lim_{x \rightarrow \infty}\frac{e^x}{{\left(1+\frac1x\right)}^{x^2}} $$ Applying L'Hospital's rule, we get the correct answer to be $\sqrt e$. However if we apply the usual rules to perform algebra on limits, i.e. limit of quotients is quotient of limits and the fact that limit of a function raised to a function is distributed over the respective functions, we can see that the denominator becomes $e^{lim_{x\rightarrow \infty} x}$ and thus the final limit becomes $1$.

I am simplifying the denominator like this $-$ $$\lim {(1+1/x)}^{x^2}=\lim {((1+1/x)^x)}^x=\lim e^{\lim x}$$

To further clarify, the limit transforms thus $-$ $$\frac {e^{\lim x}}{e^{\lim x}}=e^{\lim x-\lim x}=e^{\lim(x-x)}=e^0=1$$ where the first line follows from quotient and power rule of limits.

What am I doing wrong?

Eisenstein
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  • @AnneBauval Sorry, but I currently do not have the time to markup that much in latex. Please check the edit. – Eisenstein Jun 05 '24 at 14:06
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    The limits of both numerator and denominator are $\infty$, and you cannot conclude anything. In short: $\frac{e^\infty}{e^\infty}=e^{\infty-\infty}$ is not equal to $e^{\lim_{x\to\infty}(x-x)}=e^0=1$. It is an indeterminate form. – Anne Bauval Jun 05 '24 at 14:18
  • @AnneBauval That is indeed true, but limits can be distributed over addition or subtraction, right? As to the correct answer, I took the log and then manipulated it to apply L'Hospital's rule. – Eisenstein Jun 05 '24 at 14:22
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    "limits can be distributed over addition or subtraction": right, but not the converse. $\lim_{x\to\infty}x-\lim_{x\to\infty}x$ is not equal to $\lim_{x\to\infty}(x-x)$. It is equal to the indeterminate form $\infty-\infty$. – Anne Bauval Jun 05 '24 at 14:24
  • @AnneBauval Wow! I completely ignored that. You might want to add that as an answer. – Eisenstein Jun 05 '24 at 14:30
  • Does this answer your question? Why it is so? Please explain please help me (found using Approach$0$) – Anne Bauval Jun 05 '24 at 14:44
  • @AnneBauval Note that while many answers refer the obvious fact that $ \infty-\infty$ is not 0, none deal with the fact that lim x -lim x is not equal to zero. I think it might not be completely trivial to many beginner users of this site. – Eisenstein Jun 05 '24 at 14:58
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    @AnneBauval "it is equal to the indeterminate form ∞−∞" I think the point we both missed (or maybe I misunderstood you) is that limx-limx is not, in fact, the indeterminate form ∞−∞ (which would be the case if it were lim(x-x)). It is just not defined here, as is suggested by the top answer to the question currently linked to as duplicate. What was confusing to me was that while all the other answers I saw referred to the indeterminate form, which gives correct results once they are manipulated by l'hosptal, but here, if we consider it as an indeterminate form(which I now know is not allowed) – Eisenstein Jun 05 '24 at 16:15
  • ... we actually get the nonsensical "answer" 0. – Eisenstein Jun 05 '24 at 16:16
  • No no, again: $\lim_{x\to\infty}x-\lim_{x\to\infty}x$ (correct rewriting of your $\lim x-\lim x$ which makes no sense) is obviously the indeterminate form $\infty-\infty$, since $\lim_{x\to\infty}x=\infty$, whereas $\lim_{x\to a}(x-x)$ (correct rewriting of your $\lim(x-x)$ which makes no sense) is clearly not indeterminate (whatever $a$), as explained, precisely, in 'the top answer to the question currently linked to as duplicate':$$\lim_{x\to a}(x-x)=\lim_{x-a}0=0.$$ You seem to be still confusing both. And since $\infty-\infty$ is indeterminate, we don't get the "answer" $0$. – Anne Bauval Jun 05 '24 at 18:18
  • @AnneBauval I think you are misunderstanding me. a) The notation lim x-lim x is pretty common in many books I have read, when the context is given. B)$\lim_{x\to\infty}x-\lim_{x\to\infty}x$ is not the indeterminate form as it is just undefined, as referred to in the linked answer. The indeterminate form you are referring to can only be reproduced from $\lim_{x\to\infty}x-\lim_{x\to\infty}x$ by taking the sum inside the limit (as the correct indeterminate form is $lim{x\to a}{((f(x)-g(x))}$, where f(x) and g(x) tend to $\infty$) which is valid only for functions which have finite limits. – Eisenstein Jun 05 '24 at 18:40
  • ... .C)Yes, $\lim_{x\to a}(x-x)=0$. is indeed true, I never disputed that. – Eisenstein Jun 05 '24 at 18:40
  • Sorry, I did my best to help you. I give up. – Anne Bauval Jun 05 '24 at 18:41
  • @AnneBauval At this point I also dont know what you are referring to, as I obtained all the definitions from wikipedia (https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms). Maybe you could revisit this sometimes when you are in a better mood :) Sorry for taking up your time. – Eisenstein Jun 05 '24 at 18:47

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