I have this function
\begin{align}
\ f (x) &= (x-1)\cdot \ln\left(\frac{\ x-1}{x}\right) \\\\
\end{align}
and I need to find it's asymptotes. I know there is a vertical asymptote at $x=0$, because of the denominator in the fraction inside $\ln$. But I have a problem with finding the horizontal asymptote. I graphed it and I know it is at $y=-1$, but I don't know how to prove it. How to solve the limit of this function as $x \to \infty$ or $x \to -\infty$?.
Best Answer
Careful: the reason is that $f(x) \to \pm\infty$ as $x \to 0$, verifying this is a little more subtle than just observing the denominator $x$. Recall that $\ln x$ has a vertical asymptote at $x=0$ as well (and there is no denominator!), while your function does not have one where the logarithm's argument becomes $0$ (namely in $x=1$). You can/should check all this by calculating the corresponding limits.
Use l'Hôpital's rule: $$\lim_{x \to +\infty}\left((x-1)\ln \left(\tfrac{x-1}{x}\right)\right)=\lim_{x \rightarrow \infty}\frac{\ln \left(\tfrac{x-1}{x}\right)}{\tfrac{1}{x-1}}=\lim_{x \rightarrow \infty}\tfrac{\left(\ln \left(\tfrac{x-1}{x}\right)\right)'}{\left(\frac{1}{x-1}\right)'} =\ldots=\lim_{x \rightarrow \infty}\frac{1-x}{x}=-1$$
I omitted the calculation of the derivatives; you can do this and simplify.
The limit at $-\infty$ is similar.