how to prove n-th root of limit of f(x) is limit of n-th root of f(x)

Question

I want to know the exact proof. if $ \lim_{x\to a} f(x) = L $

$ \lim_{x \to a} \{f(x)\}^{1/n} = L^{1/n} $

most of books tell that it is special case of n squares, but I couldn't understand.

Answer 1

Lemma:

Let $a\geq 0, b \geq 0$. Then we have $$ |a^{\frac{1}{n}}-b^{\frac{1}{n}}|\leq |a-b|^{\frac{1}{n}}$$

Proof

Suppose $a\geq b$ , define $c=a^{\frac{1}{n}}$ e $d=b^{\frac{1}{n}}$, so $c-d \geq 0$ by binomial expansion we have $$c^{n}=((c-d)+d)^{n}=\sum^{n}_{k=0}{n \choose k}(c-d)^k d^{n-k}\geq d^n +(c-d)^n \geq 0 $$

so $c^n-d^n\geq (c-d)^n \geq 0$, and then $$|a-b|\geq |a^{\frac{1}{n}}-b^{\frac{1}{n}}|^{n} $$ so $$|a^{\frac{1}{n}}-b^{\frac{1}{n}}|\leq |a-b|^{\frac{1}{n}}. $$

Theorem:

If $ f(x) \geq 0$ and $\lim\limits_{x \to a} f(x)=L$ then $\lim\limits_{x \to a} (f(x))^{\frac{1}{p}}=L^{\frac{1}{p}}$, $p \in N$.

Proof:

We have that $\lim\limits_{x \to a} f(x)=L$ then $\forall \varepsilon^p >0$ we can take $\delta>0$, so that $|x-a|< \delta$ implies $|f(x)-L|<\varepsilon^p$. From this we have $|f(x)-L|^{\frac{1}{p}}<\varepsilon$, and from the previous inequality (lemma) follows $$ |f(x)^{\frac{1}{p}}-L^{\frac{1}{p}}|\leq |f(x)-L|^{\frac{1}{p}}<\varepsilon$$ then by definition of limit we conclude $\lim\limits_{x \to a} (f(x))^{\frac{1}{p}}=L^{\frac{1}{p}}$.