$\lim_{x \to 0+} {x^a}=0$ and $\lim_{x \to \infty} {x^a}=\infty$ I want to show that
when $a>0$,
$\lim_{x \to 0+} {x^a}=0$
and
$\lim_{x \to \infty} {x^a}=\infty$
I tried to find some proper $\epsilon$ and $\delta$ to prove them directly using the definition of limit but I failed.
Should I use the differentation rules for logarithmic or exponential functions?
 A: $x^a = e^{a\log(x)}.$
With $a > 0$ fixed, as $x \to 0^+, \log(x) \to -\infty.$
Therefore, as $x \to 0^+, a\log(x) \to -\infty.$
Similarly, when $x \to \infty, \log(x) \to \infty$, and therefore so does $a\log(x)$.
A: No differentiation is needed. Let $\delta<1$ be given, then the case where $\alpha\geq 1$ is obvious, hence we need only consider the case $\alpha< 1$. Since $0<\alpha<1$, there must be some $k$ such that $\frac{1}{k}<\alpha$, hence $x^{\frac{1}{k}}>x^{\alpha}$ for $x\in [0,1]$. Then, if we want $x^{\frac{1}{k}}$ to be smaller than $\delta$, we need only have $x<\delta^k$, hence putting $\epsilon=\delta^k$ satisfies the challenge.
A: (2) $f$ is said to have a limit at $\infty$ if for all $M>0$ there exists $N>0$ such that $f(x)>M$ for all $x>N$ can you take it from here?

 Let $M>0$ and $N = \sqrt[a]{M}$ $x^a > N = (\sqrt[a]{M})^a = M$


(1) We need to prove that $\forall \epsilon > 0$ $\exists \delta > 0$ such that, when $ 0 < x < 0 + \delta $, then $|f(x) - L | < \epsilon$.

 $x < \delta$ $\Rightarrow$ $|f(x) - 0| = |f(x)|< \epsilon$ taking $\epsilon = \delta$ we have the condition satisfied

