Proving the limit (delta epsilon)

[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?

[zeion]

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

[HallsofIvy]

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?
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(x[sup]2[sup]+ 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|.

[zeion]

For example, you can start by assuming that |x-1|< 1

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?