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?
 A: *

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

*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 $\ln(L)$. 
A: $e^{a-b}=\dfrac{e^a}{e^b}$, 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 $\lim\limits_{x\to\infty}f(x)=\log\left(\lim\limits_{x\to\infty}e^{f(x)}\right)$
A: Taking $x-\mathrm{Ln}(e^x+e^{-x})=y$ we have $1-e^{-2x}=e^{-y}$. So if $x\rightarrow +\infty$ then $y\rightarrow 0$. 
A: 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).
A: 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.
