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I am trying to figure out what I know to be an easy proof but I am having troubles.

Let $f:D \subset \mathbb{R} \rightarrow \mathbb{R}$ and $c$ be a limit point of $D$. If $$\lim\limits_{x \to c}\; [f(x)]^2 = 0,$$ prove that $$\lim\limits_{x \to c}\; f(x) = 0.$$

I understand this should be an easy proof but for some reason I am having trouble. One way I thought was assuming that $\lim_{x \to c} f(x)$ exists we must have that it equals zero because of the algebraic limit laws. But I feel that assuming the limit exists is using what I want to show.

Also I have looked at the fact that since $|f^2(x) - 0 | \epsilon$ whenever $|x - c| < \delta$ I know that $f(x)f(x) < \epsilon$ but I don't know how I could break that up more. Any help would be appreciated. Thank you!

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If $-\epsilon^2 \leq f(x)^2\leq \epsilon^2$, what happens if you take square roots? – Alex R. Dec 4 '12 at 4:22
Yeah, wow I can't believe I forgot that I get to pick the epsilon. All though wouldn't I need to keep it in the form $|f(x)|^2 < \epsilon^2$ to take the square root so there is no problems with $-\epsilon^2$? Thanks! – Differintegral Dec 4 '12 at 4:29
pardon, the left bound should indeed be 0 – Alex R. Dec 4 '12 at 4:30
up vote 4 down vote accepted

For any $\epsilon>0$, there is a $\delta>0$ such that when $0<|x-c|<\delta$, we have $|f(x)|^2=|f(x)^2-0|<\epsilon ^2$, so that $|f(x)-0|=|f(x)|<\epsilon$.

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This can actually be easily accomplished without an $\epsilon$ $\delta$ proof. Because $\lim_{x \to c} \frac{f(x)}{g(x)}=\frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}$, then $\lim _{x \to c} f(x)=\frac{\lim_{x \to c} f(x)^{2}}{\lim_{x \to c} f(x)}$ so that $(\lim_{x \to c} f(x))^2=\lim_{x \to c} f(x)^{2}=0$ so that $\lim_{x \to c} f(x) =0$ as desired.

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Wouldn't this be assuming that $\lim_{x \to c} f(x)$ exists before we actually know it does? Also, can we really write that ratio $\dfrac{\lim_{x \to c} f(x)^2}{\lim_{x \to c} f(x)}$ because the denominator might be zero which in this case it actually is? – Differintegral Dec 4 '12 at 5:27

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