0
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by user147263, Aaron Maroja, Community Jun 10 '15 at 1:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What does the definition of continuity say? $\endgroup$ – Tolaso Jun 10 '15 at 0:45
  • $\begingroup$ Given an $\epsilon>0$ you need to provide a corresponding $\delta>0$... and then show it works. It usually helps to work backwards. $\endgroup$ – TravisJ Jun 10 '15 at 0:46
  • $\begingroup$ $\delta$ can be chosen in terms of $\epsilon$, which is why the proof goes through in all cases. $\endgroup$ – dalastboss Jun 10 '15 at 0:59
2
$\begingroup$

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 $|x-c|<1 \leq \delta$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.