Question

lim (x^2+3x)
x->-3

I know the limit is L=0 but how do I solve further for delta!

Answer 1

since the limit is 0, you have to show you can make
\[|x^2+3x|<\epsilon\] by making
\[|x+3|<\delta\]

Answer 2

so idea is to factor what is in the absolute values signs, so that one of the terms is x + 3

Answer 3

in other words put
\[|x(x+3)|<\epsilon\] and you have control over the |x+3| term, so now you need to find some sort of bound on x

Answer 4

the trick here is to state at the outset that say
\[|x+3|<1\] which will control how big x can be. don't forget you control
\[|x+3|\] so you can make it as small as you like

Answer 5

if
\[|x+3|<1\]then
\[-1<x+3<1\] so
\[-4<x<-2\] and therefore the biggest |x| can be is 4

Answer 6

so you can pick
\[\delta = \frac{\epsilon}{4}\] and say that
\[|x^2+3x|=|x(x+3)|=|x||x+3|<\epsilon\] so long as
\[|x+3|<\frac{\epsilon}{4}\]