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?
Best Answer
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 $y = x + \cfrac{1}{e}$