Prove line asymptotic to curve I have a function denoted as:
$f(x) = \frac{x}{1+e^\frac{1}{x}}$
I want to prove the line:
$g(x)= \frac{x}{2} - \frac{1}{4}$
Is asymptotic (slant asymptote) to the above function when approaching positive infinity.
I have tried substitution variables with various expressions and various other things but I keep hitting a dead end with indeterministic limits like $\frac{0}{0}$ or $0 * \infty$.
Can some body please help with the proof, perhaps shedding some light on his approach ?
 A: Let $y=ax+b$ be a slant asyptote. Then $a=\lim_{x\to\infty}{f(x)\over x}=\lim_{x\to\infty}{1\over 1+e^{1\over x}}={1\over 2}$
Also, $b=\lim_{x\to\infty}f(x)-ax$. So $b=\lim_{x\to\infty}\frac{x}{1+e^\frac{1}{x}}-{x\over 2}=\lim_{x\to\infty}{2x-x(1+e^\frac{1}{x})\over 2 (1+e^\frac{1}{x})}=\lim_{x\to\infty}{2x-x-xe^\frac{1}{x}\over 2 (1+e^\frac{1}{x})}=\lim_{x\to\infty}{x(1-e^\frac{1}{x})\over 2 (1+e^\frac{1}{x})}=\lim_{x\to\infty}{1-e^\frac{1}{x}\over 2 (1+e^\frac{1}{x})\times{1\over x}}=-{1\over 4}$
The limit is $-{1\over 4}$ because $\lim_{t\to 0}{1-e^t\over t}=-1$.(Here $t={1\over x}$)
A: Let set $u=1/x$, then as $x \to \infty $ we get $u \to 0$.
We may write, using standard Taylor expansions:
$$
\begin{align}
\frac{x}{1+e^{1/x}}&=\frac1u \frac{1}{1+e^u}\\\\
&=\frac{e^{-u}}u \frac{1}{1+e^{-u}}\\\\
&=\frac{e^{-u}}{2u} \frac{1}{1-(1-e^{-u})/2}\\\\
&=\frac{e^{-u}}{2u}\left( 1+\frac{(1-e^{-u})}{2}+\mathcal{O}\left((1-e^{-u})^2\right)\right)\\\\
&=\frac{e^{-u}}{2u}\left( 1+\frac{1-(1-u)+\mathcal{O}\left(u^2\right)}{2}+\mathcal{O}\left(u^2\right)\right)\\\\
&=\frac{1}{2u}\left( 1-u+\mathcal{O}\left(u^2\right)\right)\left( 1+\frac{u}{2}+\mathcal{O}\left(u^2\right)\right)\\\\
&=\frac{1}{2u}-\frac{1}{4}+\mathcal{O}\left(u\right)\\\\
&=\frac{x}{2}-\frac{1}{4}+\mathcal{O}\left(\frac1x\right)
\end{align}
$$ giving the desired result.
A: Let $u = \frac{1}{x}$. Then when $x$ approaches $+\infty$, $u$ approaches $0^+$ and 
the curve is 
$$
f(u) = \frac{1}{u(1+e^u)}
$$
Meanwhile, the line (which I will call $g(x)$ to avoid the confusion of having two different $f(x)$) is
$$
g(u) = \frac{1}{2u} - \frac{1}{4}
$$
I'm going to take Laurent series; the curve and the line will be asymptotic to each other if $f(u)$ and $g(u$ agree with each other up to order $u^0$ (that is, if $f(x)-g(x) \rightarrow 0 + O\left(\frac{1}{x}\right)$)  and will not be asymptotic if $f(x)-g(x) \rightarrow C \neq 0$.  (There are possibilities between those where $f(x)-g(x)$ approaches zero but not as fast as $\frac{1}{x}$ but those cases are moot for this problem.)
$$f(u) = \frac{1}{u} \frac{1}{2+u+O(u^2)} = \frac{1}{2u} \frac{1}{1+\frac{u}{2}+O(u^2)}
=\frac{1}{2u}(1-\frac{u}{2}u+O(u^2) = \frac{1}{2u} - \frac{1}{4} + O(u)$$ 
$$g(u) = \frac{1}{2u} - \frac{1}{4}$$
The leading and constant terms of theses agree so the curve is asymptotic to that line.  
