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So far for rough work I have:
$\lvert f(x)-L\rvert=\lvert\frac{x^2+x+1}{3x+3}+1\rvert = \frac{\lvert x+2\rvert\lvert x+2\rvert}{3\lvert x+1\rvert}$
However when I take $\delta = 2$, I end up with
$0\lt\rvert x-c\lvert\lt\delta$
$\Rightarrow\rvert x+2\lvert\lt2$
$\Rightarrow-2\lt x+2\lt2$
$\Rightarrow-4\lt x\lt0$
$\Rightarrow\rvert x+1\lvert\gt-3$
$\Rightarrow\rvert x+1\lvert\geq0$ and this doesn't help me at all
I can't divide by $0$, however no matter what I choose for $\delta$, I end up with this problem because I have to - 1 from the inequality to solve for x+1 since I need to find the maximum value for the bottom in order to find $\varepsilon$.

Also, I have no idea what to do with the top. I'll end up with $\varepsilon^2$. I've never encountered a question with either of these problems before.

Katy9
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3 Answers3

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Here's the approach I taught for many years. You have gotten to $$|f(x)-L| = \frac{|x+2|}{3|x+1|}\cdot |x+2|,$$ and we want to make this expression less than a given $\epsilon>0$.

Let's stipulate first that $|x+2|<\frac12$. This means that $-\frac52<x<-\frac32$. (Note that stipulating that $|x+2|<1$ would get us in trouble, as then the denominator $|x+1|$ would get as close to $0$ as it wants.) In that interval, what is an upper bound (or if you prefer, the least upper bound) on the coefficient $\dfrac13\left|\dfrac{x+2}{x+1}\right|$? Note that $\dfrac{x+2}{x+1}=1+\dfrac1{x+1}$ and, since $-\frac32<x+1<-\frac12$, we have $-2<\dfrac1{x+1}<-\dfrac23$. Thus, $-1<1+\dfrac1{x+1}<\dfrac13$, and $\left|1+\dfrac1{x+1}\right|<1$. (You could give up on the algebra and do this graphically, of course.)

At last, set $\delta = \min(\frac12,3\epsilon)$. If $0<|x+2|<\delta$, then we have $$|f(x)-L| = \frac{|x+2|}{3|x+1|}\cdot |x+2| < \underbrace{\frac13}_{\text{using }|x+2|<1/2}\cdot |x+2| < \frac13\cdot \underbrace{3\epsilon}_{\text{using }|x+2|<3\epsilon} = \epsilon,$$ as desired.

You could do the estimate more easily, just noting that $|x+2|<\dfrac12$ and $\dfrac12<x+1<\dfrac32$. Thus, $\dfrac 1{|x+1|}< 2$, and so $\dfrac{|x+2|}{|x+1|}<\dfrac12\cdot 2 = 1$. (Because of the nature of the function and because we're dealing with negative numbers here, this coincidence isn't actually a coincidence.)

Ted Shifrin
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For $\epsilon$-$\delta$-proofs of limits for $x \to a$ it is often quite useful to replace $x = h+a$ and see what happens for $h \to 0$.

In your case:

Consider $h = x - (-2) = x+2$ and see what happens for $h \to 0$.

\begin{eqnarray*} \color{blue}{\left| \frac{x^2+x+1}{3x+3} -(-1)\right|} & = & \left| \frac{(h-2)^2+(h-2)+1}{3(h-2)+3} +1\right|\\ & = & \left| \frac{h^2 - 4h + 4+ h- 1}{3h-3} +1\right|\\ & = & \left| \frac{h^2 - 3h + 3}{3h-3} +1\right|\\ & = & \frac{1}{3}\left| \frac{h^2 - 3h + 3 + 3h-3}{h-1}\right|\\ & = & \frac{1}{3}\left| \frac{h^2}{1-h}\right|\\ & \stackrel{|1-|h|| \leq |1-h|}{\leq} & \frac{1}{3}\frac{h^2}{|1-|h||}\\ & \color{blue}{\stackrel{|h| < \frac{2}{3}}{<}} & \frac{1}{3}\frac{|h|}{|1-\frac{2}{3}|} = \color{blue}{|h|}\\ \end{eqnarray*}

Noting that $|h| = |x - (-2)|$, you find immediately that for $\epsilon >0$ you may choose $\boxed{\color{blue}{\delta = \min(\epsilon,\frac{2}{3})}}$.

trancelocation
  • 32,243
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Take $\mid x+2\mid\lt\operatorname{min}\{\sqrt\frac{3\epsilon}2,\frac12\}$. Then $\mid x+1\mid\gt\frac12$.

Now $\frac{\mid x+2\mid^2}{3\mid x+1\mid}\lt\epsilon $.