Question

Let f(x) = x^2 + 6x. Use the definition (ε − δ method) to show that
lim
x→−3 f(x) = −9

Answer 1

We remind, that by definition $\lim_{x\rightarrow a}f(x)=B$ means:

$\forall\epsilon>0$ $\exists\delta>0$: $|x-a|<\delta$ $\Rightarrow$ $|f(x)-B|<\epsilon$ .

We have to show that $\lim_{x\rightarrow -3}f(x)=-9$ . Let us fix an arbitrary $\epsilon>0$ . We need to find such $\delta>0$ that inequality $|x+3|<\delta$ will imply $|x^2+6x+9|<\epsilon$. We can rewrite the latter as: $x^2+6x+9=(x+3)^2$ From the latter it is clear that we can take $\delta=\sqrt{\epsilon}$ . Thus, $\forall\epsilon>0$ $\exists\delta=\sqrt{\epsilon}>0$ such that $|x+3|<\delta=\sqrt{\epsilon}$ implies $(x+3)^2<\epsilon$ .