Prove $\lim_{x \to c}{x^2}=c^2$ where $c$ is a real number with the $(\epsilon, \delta)$ definition. I know that you need to assume a value for $\delta$. However, I don't understand how that one assumption, which only represents one case, implies that the limit is always true. Please explain as you prove the limit.
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What does the definition of continuity say? – Tolaso Jun 10 '15 at 00:45
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Given an $\epsilon>0$ you need to provide a corresponding $\delta>0$... and then show it works. It usually helps to work backwards. – TravisJ Jun 10 '15 at 00:46
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$\delta$ can be chosen in terms of $\epsilon$, which is why the proof goes through in all cases. – dalastboss Jun 10 '15 at 00:59
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Hint: $|x^2 - c^2| = |(x+c)(x-c)|=|x+c||x-c|<|x+c|\cdot \delta \leq (|x|+|c|) \cdot \delta$
Why is $|x-c|<\delta$?
You have control to make $|x-c|<\delta$ as small as you like, can you see how to do this so you get control of how big $|x|$ can be ?
Think about for instance $|x-c|<1 \leq \delta$.
Mr.Fry
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