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My attempt: $$ \left|\frac{x^2-5x}{x^2+2}+1\right|<\left|\frac{x^2-5x}{x^2+2}\right|< \left|\frac{x^2-5x}{x^2}\right|<\frac{1}{x^2}|x^2-5x|,$$ using the restriction $|x-2|<2$, so $0<x<4$, thus $$\frac{1}{x^2}|x^2-5x|=\frac{1}{x^2}|x^2-6x+4-4+x|=\frac{1}{x^2}|(x-2)^2+x-4|<\frac{(x-2)^2}{x^2}<\epsilon,$$ hence, due to the fact that $|x-2|<\delta\le2$ we get $$ \delta^2<\epsilon x^2 < \epsilon16. $$ Finally, for any $\epsilon >0$ the corresponding $\delta$ is $\min\{\epsilon, 4\epsilon^{1/2}\}$.

Is it correct?

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  • $\begingroup$ The statement $\frac{1}{x^2}|(x-2)^2+x-4|<\frac{(x-2)^2}{x^2}$ is false. $\endgroup$
    – Aweygan
    Jul 27, 2016 at 0:54
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    $\begingroup$ Useful algebra: $\frac{x^2-5x}{x^2+2}+1=\frac{2x^2-5x+2}{x^2+2}=\frac{(x-2)(2x-1)}{x^2+2}$ $\endgroup$ Jul 27, 2016 at 0:57
  • $\begingroup$ @A.E Plugging in $x=1$ yields $2<1$ $\endgroup$
    – Aweygan
    Jul 27, 2016 at 1:03
  • $\begingroup$ Note that to get $\epsilon$-$\delta$ instead of $\epsilon-\delta$ you should use the $\epsilon$-$\delta$ construction. Also, you can use \left and \right to get properly sized delimiters (you can get even "more properly sized" delimiters using \biggl and stuff like that but I often just go with \left and \right because I am lazy. $\endgroup$ Jul 27, 2016 at 1:32
  • $\begingroup$ I like these problems...out of curiosity, what's the source? I imagine you are doing these for homework, but are you using a particular text or are they problems your instructor created (if, indeed, these problems are for a class)? $\endgroup$ Jul 27, 2016 at 1:39

1 Answer 1

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Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $\left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon$ whenever $|x-2|<\delta$. But, as Andre notes, $$ \left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|. $$ We find a positive constant $C$ such that $\left|\frac{2x-1}{x^2+1}\right|<C\Rightarrow |x-2|\left|\frac{2x-1}{x^2+1}\right|<C|x-2|$, and we can make $C|x-2|<\epsilon$ by taking $|x-2|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-2|<1$ and note the following: \begin{align} |x-2|<1&\implies 1<x<3\\[1em] &\implies 1>\frac{1}{x}>\frac{1}{3}\\[1em] &\implies 1>\frac{1}{x^2}>\frac{1}{9}\\[1em] &\implies \frac{1}{3}>\frac{1}{x^2+2}>\frac{1}{11}\\[1em] &\implies \frac{5}{3}>\frac{2x-1}{x^2+2}>\frac{1}{11}\\[1em] &\implies C=\frac{5}{3}. \end{align} Thus, we should choose $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. To see that this choice of $\delta$ works, consider the following:

Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{3\epsilon}{5}\right\}$. If $|x-2|<1$, then $\left|\frac{2x-1}{x^2+2}\right|<\frac{5}{3}$. Also, $|x-2|<\frac{3\epsilon}{5}$. Hence, $$ \left|\frac{x^2-5x}{x^2+2}+1\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|<\frac{3\epsilon}{5}\cdot\frac{5}{3}=\epsilon, $$ as desired. $\blacksquare$

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  • $\begingroup$ Wonderful answer. Thank you! $\endgroup$
    – V. Vancak
    Jul 27, 2016 at 1:29
  • $\begingroup$ You sure accepted that awfully fast. Can you guarantee it's valid? $\endgroup$
    – Aweygan
    Jul 27, 2016 at 1:32
  • $\begingroup$ @A.E Please make sure to read through it all and make sure you understand everything. There's easily a chance that I could have slipped up somewhere. $\endgroup$ Jul 27, 2016 at 1:32
  • $\begingroup$ In $5/3 > (2x-1)/(x^2+2) > 5/11$ if I plug in $x=1^+$ then I get $1/3 > 5/11$? $\endgroup$
    – V. Vancak
    Jul 27, 2016 at 1:49
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    $\begingroup$ @A.E Bingo! Yes, I was slightly careless there. Can you see how I muffed that up and how now the correction should suffice? $\endgroup$ Jul 27, 2016 at 2:01

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