# Limit of the form $\infty - \infty$

Consider: $$\lim_{x \to \infty} \left(x - \ln(e^x + e^{-x})\right)$$

I wasn't sure how to treat the $\infty - \infty$ property. Can I exponentiate the function to get $$e^x - (e^x + e^{-x}) = \frac{1}{e^x}$$ $$\lim_{x \to \infty} \frac{1}{e^x} = 0$$

I feel like I have ignored the limit part of the expression when exponentiating. Do I have to exponentiate the limit expression as well when I do this or can I ignore it for the moment?

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The exponential of $x-\ln(e^x+e^{-x})$ is not equal to $e^x - (e^x+e^{-x})$. It is equal to $\frac{e^x}{e^x+e^{-x}}$. –  Arturo Magidin Jun 26 '12 at 16:07

1. You exponentiated wrong: $$\exp(x - \ln(e^x+e^{-x})) = e^xe^{-\ln(e^x+e^{-x})} = \frac{e^x}{e^{\ln(e^x+e^{-x})}} = \frac{e^x}{e^x+e^{-x}}.$$

2. Since the exponential function is continuous, what we know is that $$\exp\left(\lim_{x\to\infty}f(x)\right) = \lim_{x\to\infty}\exp(f(x))$$ in the sense that if either one exists, then they both exist and they are equal; and that if one of them is equal to $\infty$ then they both do. Equivalently, we have that $$\lim_{x\to\infty} f(x) = \ln\left(\lim_{x\to\infty}\exp(f(x))\right).$$ So you can compute the limit of the exponential, $$\lim_{x\to\infty}\frac{e^x}{e^{x}+e^{-x}}$$ and if you obtain a value $L$, that means that the original limit will be $\ln(L)$.

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$e^{a-b}=\dfrac{e^a}{e^b}$, not $e^a-e^b$.

Otherwise, exponentiating is a good idea. The reason it helps is that the logarithmic function is continuous, so if $\lim\limits_{x\to\infty}e^{f(x)}$ exists, then $\lim\limits_{x\to\infty}f(x)=\log\left(\lim\limits_{x\to\infty}e^{f(x)}\right)$

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Why CW? $${}{}$$ –  The Chaz 2.0 Jun 26 '12 at 16:34
No particular reason why. No particular reason why not. I'm curious, why add a bunch of space to your comment with double dollar signs? –  Jonas Meyer Jun 26 '12 at 16:51
I should have just used single dollars, in retrospect. –  The Chaz 2.0 Jun 26 '12 at 17:16

Taking $x-\mathrm{Ln}(e^x+e^{-x})=y$ we have $1-e^{-2x}=e^{-y}$. So if $x\rightarrow +\infty$ then $y\rightarrow 0$.

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That's a pretty big leap from $x-\ln(e^x+e^{-x})$ to $1-e^{-2x}=e^{-y}$, given that the poster clearly needs a refresher on how exponentiation works. It seems like it should be $1+e^{-2x}$ anyway. –  Thomas Andrews Jun 26 '12 at 16:19

Replace $x$ by $\ln(x)$, apply the formula $\ln a - \ln b = \ln\frac{a}{b}$, take the limit to $\infty$ and you're done (a way without pen and paper).

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Do you have any tips on how I could improve my questions? Perhaps we should start a chat room so we do not flood the comments section here by the way. If you're willing to discuss this, feel free to join chat.stackexchange.com/rooms/3895/joe-and-chris –  Joe Jun 26 '12 at 18:37
Okay, I will keep that in mind when I ask questions from now on. –  Joe Jun 26 '12 at 18:42
When I joined the site, I recall reading posts (either from Meta or FAQ somewhere) that asked for users to display what they know about their given problem so far, where they encountered it, their work so far, where exactly they are "stuck", level of answers they should expect, avoiding imperative words in the posts such as "Show that...", et cetera. So, I try to keep all of those things in mind when writing my questions. If there is something specific I could work on, please let me know. Often I am far into a problem and get stuck on one part of it which I detail in my questions. –  Joe Jun 26 '12 at 18:47
Personally, I felt those few questions I stumbled upon yesterday were imperative, did not note where the question or idea came from, so I downvoted. Not one bit of the downvoting was personal, so I hope you don't plan on taking a personal attack of downvoting my questions out of spite. If you do though, I suppose there isn't much I can do about it. Keep in mind that I downvote many questions - currently my up/down ratio is 151/119. It has nothing to do with the users themselves, ever. It is purely based off of the question and its wording. –  Joe Jun 26 '12 at 18:52

Here's what I think most of the answers here are trying to say. First rewrite $x$ as $\ln e^x$, then use the properties of logarithms. So we have

$$\lim_{x\to\infty}[\ln e^x-\ln(e^x+e^{-x})]=$$

$$\lim_{x\to\infty}\ln\frac{e^x}{e^x+e^{-x}}=$$

$$\lim_{x\to\infty}\ln\frac1{1+e^{-2x}}$$

From here it should be easy.

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