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I'm having a little trouble figuring out how to put all my information together into an epsilon-delta proof. So far here is what I have: $f: R \setminus (-2) \rightarrow R$. Define $f(x)= \frac{x^2}{x+2}$. Show the limit as $x$ approaches 2 is equal to $f(2)$. I want to show: $|\frac{x^2}{x+2} - 1|<\epsilon$ when $0<|x-2|<\delta$. Then, $|\frac{x^2}{x+2}-\frac{(x+2)}{x+2}| = |\frac{x^2-x-2}{x+2}|=|\frac{(x-2)(x+1)}{x+2}| = |x-2||\frac{x+1}{x+2}|$. From this point on I am stuck, I am just not sure how to start the next step. Any pointers in the right direction would be of great help, as I really want to understand what I am missing and how to write my own proof in the future. Also sorry in advance for the bad editing, I was unsure of how to write a limit in LateX.

urbjhawk
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  • @Eff Thank you so much, I can't believe I didn't think of that. So here is what I have figured out now. Proof: Let $\epsilon>0$. For any $0<|x-2|<min{\frac{4\epsilon}{3},2}=\delta$, $|f(x)-f(2)|=|\frac{x^2}{x+2}-1|=|x-2||\frac{x+1}{x+2}|<\delta |\frac{x+1}{x+2}| = \delta \frac{x+1}{x+2}$ since $|x-2|<2$ $\Rightarrow$ $0<x<4$ and it follows that $\delta \frac{x+1}{x+2} < \frac{4\epsilon}{3} (\frac{2+1}{2+2}) = \frac{4\epsilon}{3} \frac{3}{4} = \epsilon$. – urbjhawk Mar 31 '17 at 02:01
  • I've made an edit in my answer responding to your comment. – Eff Mar 31 '17 at 03:54

1 Answers1

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Hint. For all $x>0$ we have that $$0<\frac{x+1}{x+2}<1$$

Responding to your comment, we can simply say the following:

Let $\varepsilon >0$ be given. Let $\delta = \min\{\varepsilon, 2\}$. From this $\delta$-definition we get $\delta\leq \varepsilon$ and $\delta \leq 2$, so that$$0<|x-2|<\delta \leq 2\quad\implies\quad 0<x<4.$$ In particular $x>0$, so we use my original hint. Then we have that $$\left|\frac{x^2}{x+2}-1\right| = \left|x-2\right|\left|\frac{x+1}{x+2}\right| <\delta\cdot1\leq\varepsilon. $$

Eff
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