Negating the Formal Definition of a Limit

Question

How to prove the $\lim_{x \to 3} (x^2) \ne 10$. Using the negation of the formal definition of a limit. I selected $\epsilon = 1$. I know I want to show that for all $\delta > 0$ that $|f(x)-L| > \epsilon$. How do I show that?

Answer 1

Well, $\epsilon=1$ is a bit too generous because the actual limit of 9 is, indeed, within 1 distance of 10. Let's pick $\epsilon=.5$. Then consider any $\delta$.

Claim: $|3-x|<\delta$ does NOT imply $|f(x)-10|<.5$.

Pick x to be $3+min(.01,\delta)$. Then $|f(x)-10| > |3.01^2-10|$ > .93 which is NOT less than our chosen $\epsilon=.5$.