3
$\begingroup$

Here's what I've done:

According to the definition, a function is continuous at $c$ if, for any $\epsilon>0$, there exists a $\delta>0$ so that, if $|x-c| < \delta$, then $|f(x)-f(c)| < \epsilon$.

$$\begin{split} |f(x)-f(c)| < \epsilon & \Leftrightarrow \left|x^{1/n}-c^{1/n}\right| < \epsilon \\ & \Leftrightarrow c^{1/n}-\epsilon < x^{1/n} < c^{1/n}+\epsilon \\ & \Leftrightarrow \left(c^{1/n}-\epsilon\right)^n < x < \left(c^{1/n}+\epsilon\right)^n, \end{split} $$ (which we can call $a < x < b$).

Thus, if we make $\delta = \min\{c-a,b-c\}$, then $$|x-c| < \delta \Rightarrow |f(x)-f(c)| < \epsilon$$ Which proves its continuity. Have I done anything wrong?

$\endgroup$

1 Answer 1

2
$\begingroup$

I think you want to take $\delta = \min\{(c^{1/n}+\epsilon)^n-c, c-(c^{1/n}-\epsilon)^n\}$ for $c\ne0$. Otherwise just take $\delta=\epsilon^n$.

EDIT: Looks like you made the correction as I was posting.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .