Proving Limits

Four step process

1. Let $ϵ>0$ be arbitrary
2. Choose $\delta$ = ?. (note that $\delta >0$)
3. Assume condition

Example

Prove ${\lim }_{x\to 2}(3x-8)=-2$ WTS: $\forall ϵ>0,\exists \delta >0$ such that $0<|x-2|,\delta ⟹|3x-8-L|<ϵ$ Let $ϵ>0$ be arbitrary Choose $\delta =\frac{ϵ}{3}$ (note that $\delta >0$) Assume $0<|x-2|<\delta$ So, $|(3x-8--2)=|3x-6|$ (by algebra) $=|3||x-2|$ (by algebra and properties of absolute value) by assumption on x, $0<|x-2|<\delta$ = $3\left(\frac{ϵ}{3}\right)=ϵ$ as required to show $\square$