TITLE: Compute $\lim\limits_{x\to 0_+}e^{\frac{1}{\sqrt{x}}}\cdot(1-\sqrt{x})^\frac{1}{x}$
QUESTION [5 upvotes]: Evaluate $$\lim_{x\to 0_+}e^{\frac{1}{\sqrt{x}}}\cdot(1-\sqrt{x})^\frac{1}{x}$$
I tried using $$\lim_{x\to 0}(1+x)^\frac{1}{x} = e$$ like so:
$$l = \lim_{x\to 0_+}e^\frac{1}{\sqrt{x}}\cdot\bigg[\big(1+(-\sqrt{x})\big)^{-\frac{1}{\sqrt{x}}}\bigg]^{\frac{-1}{\sqrt{x}}} = \lim_{x\to 0_+}e^{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}}} = e^0 = 1$$
However, the right answer is $\frac{1}{\sqrt e}$. Why is it that the whole expression in square brackets can't be taken as $e$ in this case?

REPLY [2 votes]: Considering $$A=e^{\frac{1}{\sqrt{x}}} \left(1-\sqrt{x}\right)^{\frac{1}{x}}$$ Take logarithms $$\log(A)=\frac{1}{\sqrt{x}}+\frac{1}{x}\log\left(1-\sqrt{x}\right)$$ Use Taylor series $$\log\left(1-\sqrt{x}\right)=-\sqrt{x}-\frac{x}{2}-\frac{x^{3/2}}{3}-\frac{x^2}{4}+O\left(x^{5/2}\right)$$ from which $$\log(A)=-\frac{1}{2}-\frac{\sqrt{x}}{3}-\frac{x}{4}+O\left(x^{3/2}\right)$$ Now $$A=e^{\log(A)}=\frac{1}{\sqrt{e}}-\frac{\sqrt{x}}{3 \sqrt{e}}-\frac{7 x}{36
   \sqrt{e}}+O\left(x^{3/2}\right)$$ which shows the limit and how it is approached.