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$$ \lim_{x\to 1} \frac{1}{x^2+2} = \frac{1}{3} $$

I'm having a problem proving this using $\epsilon-\delta$ proofs. For some reason When I solve for $\epsilon$, I get a negative number. Since this value is supposed to equal $\delta$ and $\delta$ can't be negative I'm not sure how to move forward. Thanks!

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Since$$\frac1{x^2+2}-\frac13=\frac{1-x^2}{3(x^2+2)}=-(x-1)\frac{x+1}{3(x^2+2)},$$the first thing to do is to choose $\delta'$ such that, when $|x-1|<\delta'$, $\left|-\frac{x+1}{3(x^2+1)}\right|$ is not large. Take $\delta'=\frac12$, for instance. Then $|x+1|<\frac52$ and $x^2+2>\frac94$. Therefore,$$\left|-\frac{x+1}{3(x^2+2)}\right|<\frac{10}{27}.$$So, take $\delta=\min\left\{\frac{27}{10}\varepsilon,\frac12\right\}$.

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  • $\begingroup$ The first equality is false. $\endgroup$ – Gribouillis Oct 11 '17 at 19:28
  • $\begingroup$ @Gribouillis Thanks. I've edited my answer. $\endgroup$ – José Carlos Santos Oct 11 '17 at 20:33
  • $\begingroup$ It is still false! $\endgroup$ – Gribouillis Oct 11 '17 at 20:49
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    $\begingroup$ @Gribouillis Not anymore… I hope! $\endgroup$ – José Carlos Santos Oct 11 '17 at 22:23
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For a given $\epsilon$, choose $\delta$ such that $$\frac{\delta(\delta+2)}{2}<\epsilon.$$ Then, if $$|x-1|<\delta$$ then $$|x|<\delta+1$$ and so $$|x+1|\leq|x|+1<\delta+2.$$

So (from the hint you received in the comments) $$\left|\frac{1}{x^2+2}-\frac{1}{3}\right|\leq\frac{|1-x^2|}{2}<\epsilon.$$

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When $x, y\in [0, 2]$, one has

$$\left|\frac{1}{x^2+2} - \frac{1}{y^2+2}\right| = \left|y-x\right| \frac{y+x}{(x^2+2)(y^2+2)}\le |y-x|\frac{4}{4} \le |y-x| $$ Take $y = 1$ and $\delta = \epsilon$.

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  • $\begingroup$ This seems more appropriate for proving uniform continuity of $f(x) =1/(x^{2}+2)$ on $[0,2]$. $\endgroup$ – Paramanand Singh Oct 11 '17 at 18:19
  • $\begingroup$ @ParamanandSingh This indeed gives a uniform $\epsilon - \delta$ proof on $[0, 2]$, but it also solves the initial question. $\endgroup$ – Gribouillis Oct 11 '17 at 19:28
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The problem is that “solving for $\epsilon$” is too restrictive in general. Let's unpack @mfl's hint. $$ \left| \frac{1}{x^2+2} - \frac{1}{3} \right| = \left|\frac{1-x^2}{x^2+2}\right| = \left|\frac{(x-1)(x+1)}{x^2+2}\right| = |x-1|\cdot |x+1| \cdot \frac{1}{x^2+2} $$ We want to make this product smaller than $\epsilon$. The only factor we can control directly is $|x-1|$; we can choose any $\delta$ we want to make that small. The trick is to control the other factors based on the assumption that $|x-1|<\delta$.

One factor is easy: $x^2 \geq 0$ for all $x$, so $x^2 + 2 \geq 2$ for all $x$, so $\frac{1}{x^2+2} \leq \frac{1}{2}$ for all $x$.

To control $|x+1|$, notice that \begin{gather*} |x-1| < \delta \implies -\delta < x-1 < \delta \\ \implies 1-\delta < x < 1+\delta \\ \implies 2-\delta < x+1 < 2+\delta \\ \end{gather*} If we know that $\delta \leq 1$, we can say $1 < x+1 < 3$, so $|x+1|<3$.
[We could use any positive number in place of $1$, but $1$ is the traditional choice.] This would give us $$ \left| \frac{1}{x^2+2} - \frac{1}{3} \right| < \delta \cdot 3 \cdot \frac{1}{2} = \frac{3}{2}\delta $$ If we know also that $\delta \leq \frac{2}{3}\epsilon$, we can conclude that the right-hand side above is $<\epsilon$, and that's exactly what we want.

We can choose $\delta$ as small as we need; how can we ensure that both $\delta \leq 1$ and $\delta \leq \frac{2}{3}\epsilon$? By setting $$ \delta = \min\left\{1,\frac{2}{3}\epsilon\right\} $$


This is not the end; you still need to write it up “forwards.” But hopefully it does explain how to work around quotients and other factors.

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For every $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that

$|x-1| \lt \delta$ implies $|\dfrac {1}{x^2+2} -1/3| \lt \epsilon$.

$F(x):=$

$|\dfrac {1}{x^2+2} -1/3| = |\dfrac {1-x^2}{x^2+2}| =$

$|\dfrac {(1-x)(1+x)}{x^2+1}|.$

Let $0\lt x \lt 2$, then $1 \lt |1+x| \lt 3,$

and $|x-1| \lt 1.$

Choose $\delta = \min ( (2/3) \epsilon, 1)$.

$|x-1| \lt \delta$ implies

$|F(x)| = |\dfrac{(1-x)(1+x)}{x^2+2}| \lt$

$|\dfrac{(1-x)(1+x)}{2}|\lt$

$(1/2)3\delta =\epsilon$.

Note: Quite straight forward, only the term $|1+x|$ needs extra attention.

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