1
$\begingroup$

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 $x \to -\infty$?.

$\endgroup$
5
  • $\begingroup$ Are you allowed to use l'Hôpital's rule? $\endgroup$ – StackTD Jan 6 '17 at 18:44
  • $\begingroup$ Yes, I think I am. Use it if you must. $\endgroup$ – Nebeski Jan 6 '17 at 18:44
  • $\begingroup$ compute the limit if $x$ tends to $+\infty$ or $x$ tends to $-\infty$ $\endgroup$ – Dr. Sonnhard Graubner Jan 6 '17 at 18:44
  • $\begingroup$ You need to do both to prove there is a horizontal asymptote $\endgroup$ – Nebeski Jan 6 '17 at 18:48
  • $\begingroup$ $ln(\frac{x-1}{x})^{x-1} = ln(\frac{x-1}{x})^x + ln\frac{x}{x-1}$ and try to compute these $\endgroup$ – juan arroyo Jan 6 '17 at 18:51
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.