Proving that $\lim_{x\to 2}\frac{x^2-5x}{x^2+2}=-1$ using the $\epsilon$-$\delta$ definition of a limit

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?

• The statement $\frac{1}{x^2}|(x-2)^2+x-4|<\frac{(x-2)^2}{x^2}$ is false. Jul 27, 2016 at 0:54
• Useful algebra: $\frac{x^2-5x}{x^2+2}+1=\frac{2x^2-5x+2}{x^2+2}=\frac{(x-2)(2x-1)}{x^2+2}$ Jul 27, 2016 at 0:57
• @A.E Plugging in $x=1$ yields $2<1$ Jul 27, 2016 at 1:03
• 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. Jul 27, 2016 at 1:32
• 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)? Jul 27, 2016 at 1:39

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. $\blacksquare$
• In $5/3 > (2x-1)/(x^2+2) > 5/11$ if I plug in $x=1^+$ then I get $1/3 > 5/11$? Jul 27, 2016 at 1:49