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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 $x \to -\infty$?.

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3 Answers 3

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I know there is a vertical asymptote at $x=0$, because of the denominator in the fraction inside $\ln$.

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.

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 $x \to -\infty$?.

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.

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$$\lim_{x\to\infty}(x-1)\log\left(\frac{x-1}x\right)=\lim_{x\to\infty}\frac{\log\left(\frac{x-1}x\right)}{\frac1x}\stackrel{l'Hos.}=\lim_{x\to\infty}\frac{\frac1{x(x-1)}}{-\frac1{x^2}}=-\lim_{x\to\infty}\frac x{x-1}=-1$$

Check now the limit when $\;x\to-\infty\;$ ...It is very similar.

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For horizontal asymptotes you have to make $x \rightarrow \infty$ and $x \rightarrow -\infty$ and $f$ must goes to some constant.

$$\lim_{x \rightarrow \infty} (x-1)\ln \left(1-\frac{1}{x}\right)=\lim_{x \rightarrow \infty}\frac{\ln \left(1-\frac{1}{x}\right)}{\frac{1}{x-1}}$$

By L'Hopital:

$$\lim_{x \rightarrow \infty}\frac{\frac{1}{x^2}\frac{x}{x-1}}{-\frac{1}{(x-1)^2}}=\lim_{x \rightarrow \infty}\frac{\frac{1}{x(x-1)}}{-\frac{1}{(x-1)^2}}=\lim_{x \rightarrow \infty}-\frac{x-1}{x}=-1$$