Csdrhrt

I am trying to prove that:

Given that the sequence: $(a_n)_{n=1}^infty$ has a limit $L$ (where $forall n in mathbb N, a_n>0$), the sequence $(a_n^r)_{n=1}^infty$ (where $r>0$, $r in mathbb{R}$) has a limit $L^r$.

I can prove in for instance for $r=frac{1}{2}$:

Given $forall epsilon > 0$ $|a_n-L|<epsilon$, one can show that $forall epsilon >0$, $$|sqrt{a_n}-sqrt{L}|=left|frac{(sqrt{a_n}-sqrt{L})(sqrt{a_n}+sqrt{L})}{(sqrt{a_n}+sqrt{L})}right|=left|frac{a_n-L}{(sqrt{a_n}+sqrt{L})}right|<epsilon.$$

But, I don't know how to make it general.
Is it true that $a_n^r-L^r leq (a_n-L)^r$,
for $a_n$, $L$ and $r$ that satisfy the above requirements?

edit: I have encountered the question in a course in which functions and continuity are thought after sequences, so my intention in the question was to use only sequences (and of course, completeness of reals, the property of Archimedes, real numbers axioms, definition of sequences limits, arithmetic sequences limits etc..)

We know that a sequence is a function whose domain is natural numbers.Let f be that function, now apply this.

What tedium.

Simply prove continuous functions preserve limits.

That is if f:R -> R is continuous and the sequence

a$_n$ -> a, then the sequence f(a$_n$) -> f(a).

To finish, note that for x > 0, x$^r$ is continuous.