Prove the following limits without using l'Hospital and Sandwich theorem Prove the following limits:
$$\lim_{x \rightarrow 0^+} x^x = 1$$
$$\lim_{x \rightarrow 0^+} x^{\frac{1}{x}}=0$$
$$\lim_{x \rightarrow \infty} x^{\frac{1}{x}}=1$$
They are not that hard using l'Hospital or the Sandwich theorem. But I curious if they can be solved with the basic knowledge of limits. I have been trying to make some famous limits like the definition of $e$ but without luck.
Thank you for your help.

Using $a^b=\exp(b\ln a)$ as definition of exponentiation with irrational exponents, it is natural to take logarithms; then the claims are equivalent to
$$\lim_{x\to 0^+}x\ln x=0,\qquad  \lim_{x\to 0^+}\frac 1x\ln x=-\infty, \qquad \lim_{x\to \infty}\frac 1x\ln x=0.$$
Substituting $x=e^{-y}$ for the first two and $x=e^y$ for the last (so that $y\to+\infty$ in all cases), they are equivalent to 
$$\lim_{y\to+\infty}\frac{-y}{e^y} =0,\qquad \lim_{y\to+\infty }(-ye^y)=-\infty,\qquad \lim_{y\to+\infty}\frac{y}{e^y}=0.$$
This makes the middle one clear and the other tow equivalent to the fact that the exponential has superpolynomial 8or at least superlinear) growth.
If not already known, this follows from the general inequality $e^t\ge 1+t$, from which find for $t\ge -1$ that $e^t=(e^{t/2})^2\ge (1+t/2)^2=1+t+\frac14t^2\ge\frac14t^2$.
