Suppose a sequence {$a_n$} such that $a_n≥0$ for all $n∈ \mathbb N$ and also $\lim\limits_{n \to ∞} a_n=L$ prove that $L\geq 0$ Suppose a sequence {$a_n$} such that  $a_n≥0$ for all n∈ N   and also $\lim\limits_{n \to ∞} a_n=L$  prove that:
(1) L ≥ 0
(2)$\lim\limits_{n \to ∞} \sqrt[m]{a_n}=\sqrt[m]{L}$
well I already prove the case when  $\lim\limits_{n \to ∞} \sqrt{a_n}$=$\sqrt{L}$ but I don't see how to prove by induction the part (2)$\lim\limits_{n \to ∞} \sqrt[m]{a_n}=\sqrt[m]{L}$
 A: For $(1),$ let $x<0.$ Then $|a_n-x|=a_n-x\geqslant-x>0$ for all $n\in\mathbb N$ so $L$ cannot be negative.
For $(2)$ use the fact that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-1}).$$ Put $x=(a_n)^{1/m},$ $y=L^{1/m}$ (by $(1)$ $L$ is nonnegative so its $m$-th root is real) and $n=m.$ Then $$\tag{0}a_n-L=\left(\sqrt[m]{a_n}-\sqrt[m]{L}\right)\left((a_n)^{\frac{m-1}{m}}+(a_n)^{\frac{m-2}{m}}L^{\frac{1}{m}}+\cdots+L^{\frac{m-1}{m}}\right).$$ If $L=0$ it is easy to conclude that $\lim\limits_{n\to\infty}\sqrt[m]{a_n}=\sqrt[m]{L}$ so you can assume that $L>0.$ Then the second factor on the right hand side of $(0)$ is positive. Letting $n\to\infty$ we see that $a_n-L\to0$ and hence $\sqrt[m]{a_n}-\sqrt[m]{L}\to0.$
A: (1) Suppose $L<0$, then $\mid a_n-L\mid>\mid \frac{L}{2}\mid$. Since $L$ is constant, this contradicts convergence. 
(2) Since the $a_n$ are positive, if they approach $0$ is has to be via positive values. Now $\sqrt[m]{a_n}$ is continuous when restricted to its domain. This means that
$$\lim_{n\rightarrow\infty}\sqrt[m]{a_n}=\sqrt[m]{\lim_{n\rightarrow\infty}a_n}=\sqrt[m]{L}.$$
