Prove that the $\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$ Can anyone share a way of proving the following limit? I'm looking for an idea that can be used at the Calc I level, using the epsilon delta formal definition.

$$\lim_{x\to a}\sqrt[n]{x}=\sqrt[n]{a}$$

Thanks.
Using the binomial theorem, you can show the following extension to the triangle inequality
$$
\left|\sqrt[n]{x_n}-\sqrt[n]{a}\right|\le\sqrt[n]{|x_n-a|}
$$
which allows a simple proof of the limit.



Proof: Assume first $a≤b$. Then expand $(\sqrt[n]a+\sqrt[n]{b-a})^n$ by the binomial theorem and only keep the two pure powers terms $a+...+(b-a)$ to get the inequality $(…)^n≥b$, or $\sqrt[n]{b-a}\ge \sqrt[n]b-\sqrt[n]a$. Then change the roles of $a$ and $b$ to obtain the other case of the absolute values.
The function $x \mapsto x^{1/n}$ is continuous everywhere.