Epsilon-Delta?

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

limx3x2=9 \large{ \lim_{x \rightarrow 3}{x^2}=9}

Definition Let f(x)f(x) be defined for all xx in some open containing the number aa with possible exception that f(x)f(x) need not to be defined at aa. We will write limxaf(x)=L\large{ \lim_{x \rightarrow a}{f(x)}=L} if given any number ϵ>0 \epsilon > 0 we can find a number δ>0 \delta > 0 such that f(x)L<ϵif0<xa<δ\large{ |f(x)-L|< \epsilon \quad \text{if} \quad 0 < |x-a|< \delta}

#Calculus

Note by Department 8
5 years, 3 months ago

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Comments

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

Given ϵ>0\epsilon>0. If ϵ1\epsilon\leq 1, let δ=ϵ12\delta=\frac{\epsilon}{12}. Now 3ϵ12<x<3+ϵ123-\frac{\epsilon}{12}<x<3+\frac{\epsilon}{12} implies 9ϵ2+ϵ2122<x2<9+ϵ2+ϵ21229-\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2}<x^2<9+\frac{\epsilon}{2}+\frac{\epsilon^2}{12^2} so that x29<ϵ|x^2-9|<\epsilon as required. If ϵ>1\epsilon>1 let δ=112\delta=\frac{1}{12}.

Otto Bretscher - 5 years, 3 months ago

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