5
$\begingroup$

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?

  • 4
    The exponential of $x-\ln(e^x+e^{-x})$ is not equal to $e^x - (e^x+e^{-x})$. It is equal to $\frac{e^x}{e^x+e^{-x}}$.2012-06-26

5 Answers 5

10
  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 $\ln(L)$.

1

$e^{a-b}=\dfrac{e^a}{e^b}$, 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 $\lim\limits_{x\to\infty}f(x)=\log\left(\lim\limits_{x\to\infty}e^{f(x)}\right)$

  • 0
    I should have just used single dollars, in retrospect.2012-06-26
1

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

  • 1
    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.2012-06-26
1

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).

  • 0
    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.2019-04-11
0

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.