Prove line asymptotic to curve

Question

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 ?

Answer 1

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 $t={1\over x}$)

Answer 2

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.

Answer 3

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.