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

Question

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.

Answer 1

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 $\frac1x \to +\infty$

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