# Asymptotes to graph

Hey I am supposed to find asymptotes of the function: $$f(x) = x \ln\left ( e+\frac{1}{x} \right )$$ The domain of the function is $$\left ( -\infty,-\frac{1}{e} \right )\cup \left ( 0,\infty \right )$$ Vertical asymptote:

I found out that limit to $$0+$$ is zero, but I do not know what to do with $$-\cfrac{1}{e}$$

Horizontal asymptote:

I got this and I am stuck:

$$\lim_{x\rightarrow \infty}\ln(e+\frac{1}{x})-x$$

Can anyone help me?

Given function $$f(x) = x \ln \bigg (e + \cfrac{1}{x} \bigg)$$ It should be obvious that there must be something bad with the function, when arguments inside the $$\ln ( \cdot)$$ goes to zero, and that is the case when $$x \to \cfrac{1}{e}$$, resulting value of $$\ln(\alpha) \to - \infty$$ as $$y \to 0$$ , where $$\alpha = e + \cfrac{1}{x}$$, which would make $$f(x) \to \infty$$

This is the only vertical asymptote, and there are no horizontal asymptotes of the function,but it has a oblique asymptote $$y = x + \cfrac{1}{e}$$

Recall the method of finding oblique asymptotes, i.e. if $$\lim_{x \to \infty} f(x) - mx = p$$ then $$y = mx + p$$ is the oblique asymptote. Here the required limit is,

\begin{align} L &= \lim_{x \to \infty} x \ln \bigg( e + \cfrac{1}{x} \bigg) - mx \\ &= x \bigg [ \ln(\cfrac{1}{em} \bigg( e + \cfrac{1}{x} \bigg) \bigg ] \\ & = x \bigg [ \ln \bigg ( \cfrac{1}{m} + \cfrac{1}{emx} \bigg) \bigg] \end{align} Which would only exist when $$m = 1$$, hence placing $$m = 1$$ $$L =\lim_{x \to \infty} x \ln \bigg( 1 + \cfrac{1}{ex} \bigg ) = \lim_{x \to \infty} \cfrac{\ln \bigg ( 1 + \cfrac{1}{ex} \bigg )}{\frac{1}{x}}$$ Which on applying L'hopital's rule $$= \lim_{x \to \infty} \bigg(1 + \cfrac{1}{ex} \bigg) \cfrac{ \cfrac{1}{ex^2} }{\cfrac{1}{x^2}} = \cfrac{1}{e}$$ Hence the equation of asymptote is $$y = x + \cfrac{1}{e}$$

• Can you explain, why is there 1/e? – siuzzy Jun 18 at 15:42
• The value of that limit, should I show how? – Ajay Mishra Jun 18 at 15:43
• Yes, I would be grateful – siuzzy Jun 18 at 15:44
• There you go. @siuzzy – Ajay Mishra Jun 18 at 15:54
• You got that, now? – Ajay Mishra Jun 18 at 15:58

Observe that $$\lim_{x\to -1/e^{-}}\ln\left(e+\frac1x\right)=-\infty$$ Then, $$\lim_{x\to -1/e^{-}}x\ln\left(e+\frac1x\right)=\left[\lim_{x\to -1/e^{-}} x\right]\left[\lim_{x\to -1/e^{-}}\ln\left(e+\frac1x\right)\right]=+\infty$$ So, $$x=-1/e$$ is an horizontal asymptote.

• why we can use L'Hopital's rule? Is it 0/0 or ∞/∞? – siuzzy Jun 18 at 15:25