# Epsilon-Delta?

Prove using "Epsilon-Delta" definition of two-sided limit:

$\large{ \lim_{x \rightarrow 3}{x^2}=9}$

Definition Let $$f(x)$$ be defined for all $$x$$ in some open containing the number $$a$$ with possible exception that $$f(x)$$ need not to be defined at $$a$$. We will write $\large{ \lim_{x \rightarrow a}{f(x)}=L}$ if given any number $$\epsilon > 0$$ we can find a number $$\delta > 0$$ such that $\large{ |f(x)-L|< \epsilon \quad \text{if} \quad 0 < |x-a|< \delta}$

Note by Department 8
2 years, 10 months ago

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Like most problems, I'm doing those in a somewhat "un-orthodox" way.

Given $$\epsilon>0$$. If $$\epsilon\leq 1$$, let $$\delta=\frac{\epsilon}{12}$$. Now $$3-\frac{\epsilon}{12}<x<3+\frac{\epsilon}{12}$$ implies $$9-\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2}<x^2<9+\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2}$$ so that $$|x^2-9|<\epsilon$$ as required. If $$\epsilon>1$$ let $$\delta=\frac{1}{12}$$.

- 2 years, 10 months ago

- 2 years, 10 months ago