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I can't remove the indeterminate form $0 \cdot \infty$. I tried to write the limit as $$\lim_{x \to 0^-}e^{x \log \left(1+\frac{1}{x} \right)}$$ but it doesn't help.

Matthew Cassell
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user515933
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    @PatrickStevens The two questions are not the same. This question has $x$ tending to $0^-$, not infinity. – Toby Mak Mar 04 '18 at 12:22
  • It isn't a duplicate. Indeed the topic of the link above has a limit to infinity and not to 0 – user515933 Mar 04 '18 at 12:22
  • It seems the hint for solving this is to write the complex number $(1+\frac{1}{x})^x$ in polar coordinates. – GEdgar Mar 04 '18 at 12:34
  • @user515933 Please remember that you can choose an aswer among the given is the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 09 '18 at 22:54

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Note that

$$\lim_{x\to 0^-}(1+\frac{1}{x})^x$$

is not well defined on reals since the base is negative and $\to -\infty$.

For $x\to 0^+$ we have

$$\lim_{x \to 0^+}e^{x \log \left(1+\frac{1}{x} \right)}=\lim_{x \to 0^+}e^{\frac{ \log \left(1+\frac{1}{x} \right)}{\frac1x}}=1$$

since for $\frac1x \to +\infty$

$$\frac{ \log \left(1+\frac{1}{x} \right)}{\frac1x}\to0$$

user
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  • It seems, as $x \to 0^-$, the imaginary part of $(1+\frac{1}{x})^x$ goes to zero and the real part goes to $1$. Using principal value for the complex exponential. – GEdgar Mar 04 '18 at 12:32
  • @GEdgar yes I mean that it is not well defined on reals, I point out this point, thanks – user Mar 04 '18 at 12:33
  • Why Wolfram alpha gives me as result the same value for the limit to $0^-$ and $0^+$? I mean, the value of limit is the same, isn't it? So why you give me two different results? – user515933 Mar 04 '18 at 12:47
  • @user515933 if you note wolfram consider also a complex part which goes to 0 thus we could say that limit is 1 but when we deal with limit, in the usual context, we consider real functions, in this sense the expression for $x\to 0^-$ in not well defined as for $lim_{x\to 0^-} \log x$, it is menaingless to be considered since the values for x are not in the domain of the function. – user Mar 04 '18 at 12:52