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.


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

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


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