# What are the asymptote(s) and hole(s), if any, of  f(x) = lnx/x?

Jul 19, 2017

The domain of the function is $D = \left(0 , + \infty\right)$
and it's continious in that domain (there are no holes).

We must calculate ${\lim}_{x \rightarrow {0}^{+}} f \left(x\right)$ so let's do it:

${\lim}_{x \rightarrow {0}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {0}^{+}} \ln \frac{x}{x} = {\lim}_{x \rightarrow {0}^{+}} \frac{1}{x} \cdot \ln x = \left(+ \infty\right) \left(- \infty\right) =$

$- \infty$

Now because ${\lim}_{x \rightarrow {0}^{+}} f \left(x\right) = - \infty$

the line $x = 0$ is a vertical asymptode of $f \left(x\right)$

Now lets calculate ${\lim}_{x \rightarrow + \infty} f \left(x\right)$

${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \ln \frac{x}{x}$

This limit is $\frac{+ \infty}{+ \infty}$ so we can use L'Hôpital's rule, so :

${\lim}_{x \rightarrow + \infty} \ln \frac{x}{x} = {\lim}_{x \rightarrow + \infty} \frac{\frac{1}{x}}{1} = {\lim}_{x \rightarrow + \infty} \frac{1}{x} = 0$

so the line $y = 0$ is a horizontal asymptode of $f \left(x\right)$