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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?

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Given function $$ f(x) = x \ln \bigg (e + \cfrac{1}{x} \bigg)$$ enter image description here 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 $y = x + \cfrac{1}{e}$

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

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

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