# A simpler epsilon-delta proof

1. Oct 18, 2009

### nietzsche

1. The problem statement, all variables and given/known data

Prove that

$$\begin{equation*} \lim_{x \to 2} x^2 + 5x -2 = 12 \end{equation*}$$

2. Relevant equations

3. The attempt at a solution

We want to prove that given $$\varepsilon > 0$$, there exists a $$\delta$$ such that

$$0<|x-2|<\delta \Rightarrow |f(x) - 12| < \varepsilon$$

$$\begin{equation*} f(x)-12\\ = x^2+5x-2-12\\ = (x+7)(x-2) \end{equation*}$$

So I have an (x-2) term there in the epsilon part. I don't know how to apply that information so that I can choose a delta. Suggestions please!

2. Oct 19, 2009

### lurflurf

write x+7 as (x-2)+9

3. Oct 19, 2009

### nietzsche

sorry, i don't follow

so when i write x+7 as (x-2)+9 i get

f(x) - 12
= (x-2)^2 + 9(x-2)

and it looks like it might be useful, but i don't know how to use it.

4. Oct 19, 2009

### aPhilosopher

can you use the fact that the the limit distributes of addition and products?

5. Oct 19, 2009

### nietzsche

i'm sorry i'm still confused. i have no idea where to go.

6. Oct 19, 2009

### nietzsche

never mind, i think i figured it out, but not with factoring it like that.

i got $$\delta = min(1,\frac{\varepsilon}{10})$$