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\rightarrow2

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|&lt; \epsilon/|x+5| but you want a constant. If \gamma&gt; |x+5| then \epsilon/\gamma&lt; \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?