If ${\lim_{x \to a}} f(x)=L$ then ${\lim_{x \to a}}\sqrt[n]{f(x)}=\sqrt[n]L$

Question

If $\displaystyle{\lim_{x \to a}} f(x)=L$ exists, then $\displaystyle{\lim_{x \to a}}\sqrt[n]{f(x)}=\sqrt[n]L$, where $n\in\mathbb{Z^+}$. Assume that $L>0$ when $n$ is even.

I have tried to prove this, but I am not sure if all the steps are right. Here's my proof:

Let $\epsilon>0$ be given such that $\epsilon<|L|$. Since $\displaystyle{\lim_{x \to a}} f(x)=L$, there exists $\delta >0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$. Now,

$$\begin{align*}|f(x)-L| &= \Big| \Big( \sqrt[n]{f(x)}\Big)^n - \big( \sqrt[n]L \big)^n \Big|\\ &= \Big| \sqrt[n]{f(x)}-\sqrt[n]L\Big| \Big|f(x)^{\frac{n-1}{n}} + f(x)^{\frac{n-2}{n}} L^{\frac{1}{n}} + \cdots + f(x)^{\frac{1}{n}} L ^{\frac{n-2}{n}} +L^{\frac{n-1}{n}}\Big|\\ & \leq \Big| \sqrt[n]{f(x)}-\sqrt[n]L\Big| \Big( \Big|f(x)\Big|^{\frac{n-1}{n}} + \Big|f(x)\Big|^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big|f(x)\Big|^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}\Big)\\ &< \epsilon\text{, whenever }0<|x-a|<\delta. \end{align*}$$

Also, $|L|\leq |L-f(x)|+|f(x)|< \epsilon +|f(x)|$, whenever $0<|x-a|<\delta$.

That is, $|f(x)|>|L|-\epsilon$, whenever $0<|x-a|<\delta.$

If $0<|x-a|<\delta$, then

$$\Big|f(x)\Big|^{\frac{n-1}{n}} + \Big|f(x)\Big|^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big|f(x)\Big|^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}>\Big(|L|-\epsilon\Big)^{\frac{n-1}{n}} + \Big(|L|-\epsilon\Big)^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big(|L|-\epsilon\Big)^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}.$$

Therefore, if $0<|x-a|<\delta$, then

$$\begin{align*}\Big| \sqrt[n]{f(x)}-\sqrt[n]L\Big| &<\frac{\epsilon}{\Big|f(x)\Big|^{\frac{n-1}{n}} + \Big|f(x)\Big|^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big|f(x)\Big|^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}}\\ &< \frac{\epsilon}{\Big(|L|-\epsilon\Big)^{\frac{n-1}{n}} + \Big(|L|-\epsilon\Big)^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big(|L|-\epsilon\Big)^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}}\\ &= \epsilon_1\text{ (say).}\end{align*}$$

Hence, $\displaystyle{\lim_{x \to a}}\sqrt[n]{f(x)}=\sqrt[n]L.$

Is the proof correct? Did I miss anything? Also, I was wondering if there is any alternative, less cumbersome way of proving this.

Answer 1

In the first equality,

\begin{align*}|f(x)-L| &= \Big| \Big( \sqrt[n]{f(x)} \Big)^n \big( \sqrt[n]L \big)^n \Big|\\ &= \Big| \sqrt[n]{f(x)}-\sqrt[n]L\Big| \Big|f(x)^{\frac{n-1}{n}} + f(x)^{\frac{n-2}{n}} L^{\frac{1}{n}} + \cdots + f(x)^{\frac{1}{n}} L ^{\frac{n-2}{n}} +L^{\frac{n-1}{n}}\Big|\\ & \leq \Big| \sqrt[n]{f(x)}-\sqrt[n]L\Big| \Big( \Big|f(x)\Big|^{\frac{n-1}{n}} + \Big|f(x)\Big|^{\frac{n-2}{n}} |L|^{\frac{1}{n}} + \cdots + \Big|f(x)\Big|^{\frac{1}{n}} |L| ^{\frac{n-2}{n}} +|L|^{\frac{n-1}{n}}\Big)\\ &< \epsilon\text{, whenever }0<|x-a|<\delta. \end{align*}

You've missed a minus sign. It should read $$\Big| \Big( \sqrt[n]{f(x)} \Big)^n -\big( \sqrt[n]L \big)^n \Big|$$ instead.

To be rigorous, You need $\epsilon < |L|$, so that $(|L|-\epsilon)^{1/n}$ is well defined. At the beginning of your proof, you've only assum that $\epsilon > 0$.

Finally, $\epsilon_1$ is not properly defined.

To simplify the argument, you may consider the sequential criterion for limits.