$$L:=\lim_{x\to0^{+}}x^{a}=0~~~~~~~~~~~\left(a>0\right)$$
As$~a\geq1~$, it is quite obvious for me that the above limit converges to zero.
The current problem for me is that$~a<1~$
For instance as we fix$~a=10^{-4}~$
$$x^a=x^{10^{-4}}=x^{\frac{1}{10^{4}}}$$
$$=\sqrt[10^{4}]{x}$$
$$f\left(x\right):=\sqrt[10^{4}]{x}$$
What I want to prove is that for$~0<x_{\text{small}}<x_{\text{big}}~$, the following is satisfied.
$$\color{blue}{\sqrt[10^{4}]{x_{\text{small}}}<\sqrt[10^{4}]{x_{\text{big}}}}$$
For simplicity, I assume$~x_{\text{small}},x_{\text{big}}=10^{-9},10^{-3}~$respectively.
So, of course,$~f\left(x\right)~$indicates the value of piece which is powered by$~10^{4}~$is$~x~$
$$\sqrt[10^{4}]{10^{-9}},\sqrt[10^{4}]{10^{-3}}$$
$$\text{Assumption}~\rightarrow~10^{-t}:=\sqrt[10^{4}]{10^{-9}}$$
$$\text{Assumption}~\rightarrow~10^{-u}:=\sqrt[10^{4}]{10^{-3}}$$
$$10^{-t}=\left(10^{-9}\right)^{\frac{1}{10^{4}}}$$
$$=10^{-\frac{9}{10^{4}}}$$
$$\therefore~~t=\frac{9}{10^{4}}~,~u=\frac{3}{10^{4}}$$
\begin{equation*}%uasge:&smth\\. . Dont write symbol of line break at the end of row \begin{cases} 10^{-t}=10^{-\frac{9}{10^{4}}}\\ 10^{-u}=10^{-\frac{3}{10^{4}}}\\ \end{cases} \end{equation*}
My brain has panicked at here.
How can I proceed?
For any fixed positive $a$ the function given by $$ f(x) = x^a $$ is a continuous increasing function of $x$, with $f(0) = 0$. One way to see that is to note that its derivative $$ f'(x) = ax^{a-1} > 0 $$ for $x > 0$.
The fact that $a-1 < 0$ when $a < 1$ means that the graph of $f$ is tangent (from the right) to the $y$-axis at the origin. So the slope is badly behaved there, but the function itself has limit $0$.
Look at the graph of $y = \sqrt{x}$ to see what is going on.