$\lim_{x\rightarrow 0^+}x^x$

Question

How can I calculate $\lim_{x\rightarrow 0^+}x^x$?

I can only write it in the form $e^{x\ln x}$. I would like to use L'Hospital rule somehow, but I can't write it in form of fractions.

Answer 1

HINT:

$$y=x^x\iff \ln y=x\ln x=\frac{\ln x}{\frac1x}$$ which is of the form $\frac\infty\infty$ as $x\to 0^+$

So applying L'Hospital's Rule

$$\lim_{x\to0^+} \ln y=\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac1x}=\lim_{x\to0^+}\frac{\frac1x}{-\frac1{x^2}}=\lim_{x\to0^+}(-x)=0$$

Answer 2

Consider $f(x) = x \ln x$ separately. Note that $\lim_{t \to \infty} e^{-t} = 0$, so we have $\lim_{x \downarrow 0} f(x) = \lim_{t \to \infty} f(e^{-t})= \lim_{t \to \infty} e^{-t} (-t) = - \lim_{t \to \infty} \frac{t}{e^t} = 0$.

Hence it follows that $\lim_{x \downarrow 0} x^x = \lim_{x \downarrow 0} e^{f(x)} = 1$.

Answer 3

$\lim_{x\rightarrow 0^+}x^x= e^{\lim_{x\rightarrow 0^+}x\ln x}$as $e^x$ is a continuous function.

$\lim_{x\rightarrow 0^+}x\ln x=\lim_{x\rightarrow 0^+}\frac{\ln x}{\frac{1}{x}}$

Now,you can apply L'Hopital's Rule