0
$\begingroup$

I need to compute the limit of this expression: $$\lim_{x\to 0} \frac{a^x - b^x}{cx^3 + dx^2}$$. In the solution (given, in the link) they used L'Hospital twice. I understand the first time, but in the second time I can't see why (and how could they) use it. i.e the conditions weren't there... am I missing something?

solution to the problem

$\endgroup$
  • 2
    $\begingroup$ You are right to question the result. The solution you were shown is incorrect. $\endgroup$ – Umberto P. Feb 7 '18 at 18:36
  • 2
    $\begingroup$ You can't use the Hospital twice on this one. $\endgroup$ – Lord Shark the Unknown Feb 7 '18 at 18:36
  • $\begingroup$ Any suggestions how to compute it otherwise? $\endgroup$ – Daniel Feb 7 '18 at 20:27
2
$\begingroup$

Good job! You are absolutely correct to question this solution.

In the second limit, the limit of the denominator is $$\lim_{x \rightarrow 0} \left(3cx^2 + 2dx\right) = 0,$$ while the limit of the numerator is $$\lim_{x \rightarrow 0} \left(\ln(b) b^x - \ln(a) a^x\right) = \ln(b) - \ln(a),$$ which is not zero unless $a = b$. This means that the second application of l'Hôpital's rule in the solution is invalid.

Even seasoned instructors sometimes make mistakes when preparing examples or solutions. It's good to verify things for yourself, as you have wisely done in this case.

$\endgroup$
  • $\begingroup$ Wasnt sure about this thought, but after you said it... thanks $\endgroup$ – Daniel Feb 7 '18 at 20:23

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.