Proving the limit (delta epsilon)

[zeion]

Homework Statement

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}

Homework Equations

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]

Homework Statement

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}

Homework Equations

The Attempt at a Solution

0 &lt; |x - 1| &lt; \delta \rightarrow 0 &lt; |\frac{x + 3}{x^2 + x + 4} - \frac{2}{3}| &lt; \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+ 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&lt;\epsilon so you need |x-1|&lt; 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|.</sup>

 

[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?