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Let $f(x) = \sqrt{x}$. I'm trying to show that $f$ is continuous at $0$ by contradiction.

In order for a function to not be continuous at $0$: There exists an $\epsilon > 0$ such that for every $\delta_n = \frac{1}{n} > 0$, $|x_{n} - 0| \leq \delta_n \implies |\sqrt{x_n} - 0| > \epsilon$ where $x_n$ is an element in the domain of $f$.

Suppose $\epsilon = 1$. Then I see that $|\sqrt{x_n}| > 1 \implies x_n > 1$. But I see that $x_n \to 0$ since $\delta_n = \frac{1}{n} \to 0$.

This would be a contradiction because $x_n > 1$ which means that $\{x_n\}$ will never converge to 0.

Could I get some feedback on my proof?

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    You've only shown that that there can't be such an $\epsilon$ in $[1, \infty)$. You also need to show that there is no such $\epsilon$ in $(0,1)$. The proof can be carried for all cases at once by looking at the first $\delta_n$ such that \delta_n < \epsilon^2 (which exists by the Archimedean property).2012-04-11

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If you assume that $f$ is not continuous at 0, here's what you would know:

There exists an $\epsilon>0$ such that for every $\delta_n={1\over n}$, there exists an $x_n$ such that $0\le x_n<\delta_n$ and $\sqrt{x_n}\ge\epsilon$.

You do not get to choose this $\epsilon$; in particular, you do not know it is equal to $1$. So your argument falls apart here. But you can say "Let $\epsilon>0$ be a number satisfying the condition of the preceding paragraph". You don't know the actual value of $\epsilon$, but you do know it is a fixed, positive number. Then, towards deriving a contradiction, what you could do is find a $\delta_n$ such that no number $x_n$ satisfies both $0\le x_n<\delta_n$ and $\sqrt{x_n}\ge\epsilon$. (Towards this end, consider the square of $\epsilon$.)

Incidentally, a direct proof would be much neater in my opinion (note that in the suggested argument above, you are essentially proving continuity at 0 directly).