Epsilon-Delta Linear Case

Remember the Limits Formalized Definition $\forall ϵ>0,\exists \delta >0suchthat0<|x-c|<\delta ⟹|f(x)-L|<ϵ$ If we have a limit like: ${\lim }_{x\to 3}(-2x+1)=-5$ Then:

• $c=3$
• $L=-5$
• $f(x)=-2x+1$
• $x=x$

Rough Work

We would then find a value $\delta$ such that $|x-3|<\delta$, $|-2x+1-(-5)|<ϵ$ Rearrange the epsilon statement $|-2x+6|<ϵ$ 2|x-3|<\epsilon\ $|x-3|<\frac{ϵ}{2}$ Then, we can say that $\delta =\frac{ϵ}{2}$

Proof

Let $ϵ>0$ be arbitrary Choose $\delta =\frac{ϵ}{2}$ Assume $|x-3|<\delta$ Assume $|-2x+1-(-5)|<ϵ$ Now, $|-2x+1-(-5)|=|-2x+6|=2|x-3|<2\frac{ϵ}{2}$ $2|x-3|<ϵ$ Thus, $|(-2x+1)-(-5)|<ϵ$ $\square$