Okay, rewrite to $$e^{\ln(2-a^{1/x})\cdot x}$$ Now because of the continuity of the exponential function we look at $\ln(2-a^\frac{1}{x})\cdot x$, but here's the "problem".

$\ln(2-a^\frac{1}{x})$ tends to $\ln(1)=0$ but the limit for $x$ obviously does not exist. Therefore the limit theorems are not applicable, because they require each limit to exist on its own.

Now colloquially speaking $\ln(2-a^\frac{1}{x})$ should make everything zero, and $x$ should make everything infinity as the limit approaches infinity. I know from Wolfram Alpha, that the complete limit of both functions "equals" infinity. But I don't know how to show it. Maybe you could use L'hopital but I do not see how. How does one approach such a situation?

EDIT: Don't know if it is relevant, but $ 0 < a < 1 $ and $x \in \mathbb{R}_{+} $ without zero.


Hint (with, of course, $\;a>0\,$ ):

$$\lim_{x\to\infty}\frac{\log(2-a^{1/x})}{\frac1x}\stackrel{\text{l'Hospital}}=\lim_{x\to\infty}\frac{\frac{\frac1{x^2}a^{1/x}\log a}{2-a^{1/x}}}{-\frac1{x^2}}=\lim_{x\to\infty}-\frac{a^{1/x}\log a}{2-a^{1/x}}=-\frac{1\cdot\log a}{2-1}$$

  • $\begingroup$ Very nice, I think I need some further arithmetic algebra practice to be able to see these tricks $\endgroup$ – Jonathan Apr 1 '17 at 12:32
  • $\begingroup$ @PrayasAgrawal I can't see why you think that: $\;2-a^{1/x}\xrightarrow[x\to\infty]{}2-1=1\;$ ... $\endgroup$ – DonAntonio Apr 1 '17 at 12:33
  • $\begingroup$ actually Prayas the L'hopital's rule can be extended to various indefinite limit types such as $0 \cdot \infty $ $\endgroup$ – Jonathan Apr 1 '17 at 12:33
  • $\begingroup$ @PrayasAgrawal Did you read my comment addressing yours? Do you understand what I am trying to tell you there? $\endgroup$ – DonAntonio Apr 1 '17 at 12:35
  • $\begingroup$ @PrayasAgrawal Because $\;a^{1/x}\;$ continuous everywhere at $\;\Bbb R\setminus\{0\}\;$ , and at the limit it is like $\;a^0=1\;$ ... $\endgroup$ – DonAntonio Apr 1 '17 at 12:36

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.