Proof of limits, Spivak

Question

So when proving that the function $f(x) = \frac 1x$ approaches $\frac 13$ as $x \to 3$ , Spivak does something like:

$$|\frac 1x - \frac 13| < \epsilon$$

he simplifies it to

$$\frac 13 \cdot \frac1{|x|} \cdot |x - 3| $$

so basically he is trying to show that $(x-3)$ is small hence $\epsilon$ is bigger, so $x$ in $x-3$ lies between $2$ and $4$ i.e. $2<x<4 $ and $\frac 1x $ lies between $\frac 14 $ and $\frac 12$ ie. $\frac14 <\frac 1x<\frac 12$

I understand all the way till there however,

he then puts an inequality:

$$\frac 13 \cdot \frac 1{|x|} \cdot |x-3| < \frac 16 |x - 3|$$

I was just wondering if anyone can inform me on where the $\frac 16$ came from?

score 1 · Accepted Answer · answered May 24 '14 at 19:13

1

We're letting $x\to 3$. Since $x$ is close to $3$, we can assume that $|x-3|<1$. Since $||x|-|3||<|x-3|$ we get that $||x|-3|<1$ i.e. $-1<|x|-3<1$. Adding three gives $2<|x|<4$. Inverting gives $$\frac{1}{|x|}<\frac 1 2$$

answered May 24 '14 at 19:13

Pedro

122,002

that's helpful thanks, but where did he get the 1/6 from? – LucilVerma May 24 '14 at 20:19
@Wangzmk What $2$ times $3$? (Also, you might have mistakenly downvoted instead of upvoting.) – Pedro May 24 '14 at 20:21
Oh okay got it. (I didn't down vote, I don't have enough reputation to even up vote.) Thanks – LucilVerma May 24 '14 at 21:21

Proof of limits, Spivak

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