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.

METHODOLOGY $1$
Here is one way to evaluate the limit without use of L'Hospital's Rule.  We make use of the inequality 
$$e^x\ge \left(1+\frac xn\right)^n \tag 1$$
for $x>-n$, which I proved in THIS ANSWER using only the limit definition of the exponential function along with Bernoulli's Inequality.  Note that setting $n=2$ in $(1)$ reveals 
$$e^x\ge 1+x+\frac14x^2 \tag 2$$
for $x>-2$.
Proceeding with the evaluation of the limit $\lim_{x\to 0^+}x^x$, we note that we can write $x^x=e^{x\log(x)}$.  Next, we have 
$$\begin{align}
\lim_{x\to 0^+}x\log(x)&=-\lim_{x\to \infty}xe^{-x}\\\\
&=-\lim_{x\to \infty}\frac{x}{e^x}
\end{align}$$
From $(2)$, we find that the term $\frac{x}{e^x}$ satisfies the inequalities
$$0\le \frac{x}{e^x}\le \frac{x}{1+x+\frac14x^2} \tag 3$$
whereupon applying the squeeze theorem to $(3)$ reveals
$$\lim_{x\to \infty}\frac{x}{e^x}=0$$
Finally, since $e^0=1$, we find the limit of interest is indeed
$$\lim_{x\to 0+}x^x=1$$ 

METHODOLOGY $2$
Alternatively, I showed in THIS ANSWER and THIS ONE, using only the limit definition of the exponential function along with Bernoulli's Inequality, that the logarithm function satisfies the inequalities
$$\frac{z-1}{z}\le \log(z)\le z-1 \tag 4$$
Setting $z=x^\mu$, $\mu<1$ in $(4)$ yields
$$\frac{x^\mu -1}{\mu x^\mu}\le \log(x)\le \frac{x^\mu -1}{\mu}$$
Therefore, we have 
$$\frac{x(1-x^{-\mu})}{\mu}\le x\log(x)\le \frac{x(x^\mu -1)}{\mu} \tag 5$$
whereupon applying the squeeze theorem to $(5)$ yields
$$\lim_{x\to 0^+}x\log(x)=0$$ 
and therefore
$$\lim_{x\to 0^+}x^x$$
