# Proving the limit (delta epsilon)

1. Oct 8, 2009

### zeion

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

Prove the follow statements directly using the formal $$\epsilon , \delta$$ definition.

$$\lim_{x\rightarrow 1} \frac{x + 3}{x^2 + x + 4} = \frac{2}{3}$$

2. Relevant equations

3. The attempt at a solution

$$0 < |x - 1| < \delta \rightarrow 0 < |\frac{x + 3}{x^2 + x + 4} - \frac{2}{3}| < \epsilon$$

Not sure what to do now.

0 < | 3(x+3) - 2(x2 + x + 4) / 3(x2+x+4) | < e
0 < | -2x2 + x + 1 / 3(x2+x+4) | < e
0 < | (-2x - 1)(x - 1) / 3(x2+x+4) | < e

Now I can control (x - 1), but how do I do this?

2. Oct 9, 2009

### zeion

Should I have posted this in the "Calculus and Beyond" section? :/

3. Oct 9, 2009

### HallsofIvy

Now you need to get a bound on $|(2x+1)/3(x^2+ x+ 4)|$

For example, you can start by assuming that |x-1|< 1 so that 0< x< 2. Then 0< 2x< 4 and 1< 2x+1< 5. Also 0< x2< 4 so 0< x2+ x< 6, 4< x2+ x+ 4< 10 and 12< 3(x2+ x+ 4)< 30. That tells you that 1/30< 1/3(x2+ x+ 4)< 1/12 and so that 1/30< (2x+1)/3(x2+ x+ 4)< 5/12.

Now you know that |x-1|/30< (2x-1)|x-1|/3(x2+ x+ 4)< 5|x-1|/12

Since you want to make sure that is less than $\epsilon$, you want $5|x-1|/12<\epsilon$ so you need $|x-1|< 12\epsilon/5$. To make certain that |x-1|< 1 so all of that is true, take $\delta$ to be the smaller of 1 and $12\epsilon/5|$.

4. Oct 14, 2009

### zeion

I've been told that I should only pick values that are less than whatever x is approaching (ie. <1 in this case), is this true?