Solving limit shape $0^0$ I have an exercise which is $\lim_{x \to 0^+} x^x$.
I have done it, and the answer is $1$, but I just need to know the easiest way to solve it without l'Hospital's rule, since in my country they don't allow us to use L'Hospital's rule.

Note that for $0 < x < 1$ we have the inequality $1-x < -\ln x < (1-x)/x.$
Hence,
$$2x (1- \sqrt{x}) < -x\ln x = -2x\ln \sqrt{x} < \frac{2x(1-\sqrt{x})}{\sqrt{x}}.$$
Applying the squeeze theorem we have $\lim_{x \to 0} x \ln x = 0$ and using the continuity of the exponential function
$$\lim_{x \to 0}x^x = \lim_{x \to 0}\exp(\ln x^x)=\lim_{x \to 0}\exp(x\ln x)=\exp(\lim_{x \to 0}x \ln x)= 1.$$
