Limiting Proof: Showing lim (x^2+3x) = 10 x→2

[roadworx]

Homework Statement

Use the precise definition to show
$$lim (x^2+3x) = 10$$
x$$\rightarrow$$2

The Attempt at a Solution

Let $$\epsilon$$ > 0

$$x^2 + 3x - 10 < \epsilon$$

$$(x-2)^2 = x^2 - 4x + 4$$

This doesn't equal the equation. Add 7x, -14

$$\left| x-2 \right| ^2 + 7 \left| x-2 \right|$$

So far it's alright. Now I need to get a value for $$\delta$$

$$\epsilon$$ = $$\delta^2 + 7 \delta$$

Now I'm totally confused. Normally I've used simply $$\delta$$ expressions like $$\delta$$ = $$\epsilon$$/2. What should I say my $$\delta$$ is equal to in this case, and why?

So $$\left| (x^2 + 3x) -10 \right|$$ < $$\delta^2 + 7 \delta$$

Any help?

 

[HallsofIvy]

It would be better to use the fact that x²+ 3x- 10= (x- 2)(x+ 5) so
|x²+ 3x- 10|= |x-2||x+ 5| and you want that less than $\epsilon$. That will be true if $|x-2|< \epsilon/|x+5|$ but you want a constant. If $\gamma> |x+5|$ then $\epsilon/\gamma< \epsilon/|x+5|$ so you can set $\delta= \epsilon/\gamma$. To find an upper bound on |x+ 5| remember that you are taking x close to 2 anyway: say |x- 2|< 1 so -1< x-2< 1. How large can x+ 5 be?