$$\lim_{x \to 0}x^x = 1$$
Some sources say that this is solvable L'hopital rule, and I am unsure how I can use the rule to prove this. Can anyones show me how?
$$\lim_{x \to 0}x^x = 1$$
Some sources say that this is solvable L'hopital rule, and I am unsure how I can use the rule to prove this. Can anyones show me how?
$x^x = e^{x\ln x}$, so $$\lim_{x\to 0}x^x=\lim_{x\to 0}e^{x\ln x}=e^0=1$$.
We can rewrite $x^x=e^{x\ln x}$, now using continuity of exponentiation we know that $$\lim_{x\to 0}e^{x\ln x}=e^{\lim_{x\to 0} x\ln x}$$
Calculating $\lim\limits_{x\to0} x\ln x$ is simpler, and it is indeed $0$ (you can use L'Hospital to prove this limit), now we have: $$\lim_{x\to 0}x^x=\lim_{x\to0} e^{x\ln x}=e^0=1.$$