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Choose $\delta = \min (1, \frac{\epsilon}{10})$ is the following statement true?

$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$

Okay so this is what I have so far

$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}{|x^2+1|}$$

I then chose $\delta = \frac{1}{2}$ (because if I chose $\delta = 1$, $x$ would come out as $2$ which would give me $|x-1|\cdot 0 < \epsilon$ which doesn't tell me much?)

So if $\delta = \frac{1}{2}$, then $\frac{1}{2}< x < \frac{3}{2}$ (because it's a fraction I used $x > \frac{1}{2}$)

$$|x-1|\cdot\frac{\left|-\frac{1}{2}+2\right|}{\left|(\frac{1}{2})^2+2\right|}=\frac{2}{3}|x-1|< \epsilon$$ $$|x-1|< \frac{3}{2}\epsilon$$

I'm just so confused because I don't know how to relate the value for $\delta$ (which I found to be $\frac{3}{2}\epsilon$ to the delta they're making me choose of $\min (1, \frac{\epsilon}{9})$. I'm also not sure if I chose the correct value for $x$ as it is a fraction.

Sil
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Haley
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    Mathematical formulae look better in $\LaTeX$. Here is a quick tutorial. – Τίμων Apr 16 '16 at 09:36
  • Hint: If $|x-1|<\delta$ what do you know about $|2-x|=|x-2|$ and $|x^2+1|$? – Christian Apr 16 '16 at 09:42
  • does that mean δ|x-2|/|x^2+1| < ϵ? and then would I let δ = 1 so |x-2|/|x^2+1| < ϵ – Haley Apr 16 '16 at 09:58
  • @Laylah I think you need to get a bound in terms of $\delta$ for $|x-2|$ and $|x^2+1|$. Note that $\delta=1$ only works if $\varepsilon \geq 10$ since $\delta\leq \frac{\varepsilon}{10}$. – Christian Apr 16 '16 at 10:02
  • Do you mean like get an individual bound for each? So if I let δ = 1/2 then |x-1|<1/2 and |x-2| <7/2 and then 5/4<|x^2+1|<13/4 so 4/13<1/|x^2+1|<4/5 – Haley Apr 16 '16 at 10:17
  • @Laylah yes but without replacing $\delta$ by a fixed number, since $\delta\leq \varepsilon/10$ und you did not fix $\varepsilon$, e.g. $|x-2|=|x-1-1|\leq |x-1|+1\leq \delta +1$. – Christian Apr 16 '16 at 10:20
  • Where on earth are all these $\delta=\min(\epsilon,1/10)$ questions coming from? Cf. http://math.stackexchange.com/questions/1743400/show-that-delta-min-left-1-frac-epsilon10-right-implies-lim-x-to-1. – Hans Lundmark Apr 16 '16 at 10:26
  • @user93559 So I do the same thing, but use δ≤ε/10 ? does that mean I have |x-1|<ε/10 so 10|x-1| <ε and 10|x-2|/|x^2+1| < ε^2? – Haley Apr 16 '16 at 10:36
  • Definitely don't do the same thing. That's a completely different function – K.Power Apr 16 '16 at 10:38
  • I'm so confused – Haley Apr 16 '16 at 10:41
  • @Laylah What I meant is: for the moment just keep the $\delta$ and find a upper bound for $|x-2|$ (as above) and an upper bound for $1/|x^2+1|$. – Christian Apr 16 '16 at 10:42
  • But how can I find an upper bound if I don't choose a delta, because can't x take on any value for either |x−2| and 1/|x^2+1|? Unless I can say that 1/|x^2+1| at most must be 1 because the smallest value x^2 can take is 0? – Haley Apr 16 '16 at 10:48
  • @Laylah: I posted an answer. Concerning your question: since $x^2\geq 0$, we get that $|x^2+1|=x^2+1\geq 1$ and hence $1/|x^2+|\leq 1$. And by triangle inequality we get $|x-2|\leq |x-1|+1\leq \delta +1$. In other words if $x$ is $\delta$ close to $1$ it has to be $1+\delta$ close to $2$. – Christian Apr 16 '16 at 10:54
  • @Laylah By the way, your above argument on $1/|x^2+1|\leq 1$ is perfectly fine - it is just a different formulation of what I wrote my comment. – Christian Apr 16 '16 at 10:56
  • @user93559 thank you, that explanation really helped. Especially for the bounds on |x-2| :) – Haley Apr 16 '16 at 11:00
  • @Laylah Just out of curiosity: why did you start with $\delta\leq \min{1,\varepsilon/10}$? It turns out $\delta\leq \min{1,\varepsilon/2}$ would be enough. – Christian Apr 16 '16 at 11:01
  • @user93559 the question we were given asked us to prove or disprove that statement given that δ</={1,ε/10} so that's why. I think it made the question harder – Haley Apr 16 '16 at 11:06

2 Answers2

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An alternative answer keeping the $\delta$: We know that $|x-1|<\delta$ and $$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}$$ Now, we use the bounds $$|x-2| = |x-1-1|\leq |x-1|+1 \leq\delta +1$$ and $$|x^2+1|=x^2+1\geq 1$$ to obtain $$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}\leq \delta(1+\delta) \leq \delta2 \leq \frac{\varepsilon}{5}<\varepsilon$$ using $\delta\leq \min\{1,\frac{\varepsilon}{10}\}$.

Christian
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As $|x-1|<1$ for all possible $\delta$ values then $$|x-2|< 2\ \text{and} \ |x^2+1|>1$$ So then we have $$|x-1|\frac{|-x+2|}{|x^2+1|}< \frac{2|x-1|}{1} = 2|x-1|$$

If $\epsilon\geq10$ then remembering $|x-1|<1$ as $\delta =1$ in this case $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot 1<10\leq\epsilon$$ as desired. If $\epsilon<10$ then remembering $|x-1|<\delta=\frac{\epsilon}{10}$ we have $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot \frac{\epsilon}{10}=\epsilon/5<\epsilon$$ and you are finished. Let me know if you have questions.

K.Power
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