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Show that $f(x) = \sqrt{1-x^2}$ is uniformly continuous on $[-1,1]$.

I showed it was differentiable on the open interval (-1,1), which implies continuous on $(-1,1)$. However, I still need to show continuity at $1$ and $-1$ so I can use the fact that continuous on a closed interval implies uniform continuity.

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    $\begingroup$ Do you know that $\sqrt x$ is continuous on $[0,\infty)$? $\endgroup$ – user99914 Sep 8 '15 at 0:35
  • $\begingroup$ Yes, however i'm not sure I see how to apply it here. $\endgroup$ – 1233211 Sep 8 '15 at 1:00
  • $\begingroup$ To prove the continuity at $-1$ and $1$, just show the limit is the same from both ends. I.e. Show that $lim_{x->1+} f(x) = lim_{x->1-} f(x)$ $\endgroup$ – asosnovsky Sep 26 '15 at 14:55
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Note that $\sqrt x$ is continuous on $[0, \infty)$. Thus $\sqrt{1-x}$ and $\sqrt{1+x}$ are continuous on $[-1, 1]$. Hence

$$\sqrt{1-x^2} = \sqrt {1-x}\sqrt{1+x}$$

is also continuous on $[-1, 1]$.

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