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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.

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

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

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  • $\begingroup$ 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

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