Question

Use the epsilon delta limit definition to show that

Answer 1

\(\Huge \lim_{x \rightarrow 3}\frac{1}{x}=\frac{1}{3}\)

Answer 2

I'm having trouble turning the value
\(|x-3|\) into something that resembles \(|\frac{1}{x}-3|\) enough to proceed with the limit definition

Answer 3

I read the example at http://math.stackexchange.com/questions/418961/epsilon-delta-proof-that-lim-limits-x-to-1-frac1x-1 , but it doesn't seem to immediately apply to this case...

Answer 4

You have to establish the following:\[0<\left|x-3\right|<\delta~~\Rightarrow~~\left|\frac{1}{x}-\frac{1}{3}\right|<\epsilon\]
With some preliminary analysis, you can find
\[\left|\frac{1}{x}-\frac{1}{3}\right|=\left|\frac{3}{3x}-\frac{x}{3x}\right|=\frac{1}{3}\frac{1}{|x|}|x-3|\]
Now you can try finding an appropriate \(\delta\).

Answer 5

I would suggest assuming \(\delta\le1\), then
\[|x-3|<1~~\Rightarrow~~2<x<4~~\Rightarrow~~\frac{1}{4}<\frac{1}{x}<\frac{1}{2}~~\Rightarrow~~\frac{1}{|x|}<\frac{1}{2}\]
So now
\[\frac{1}{3}\frac{1}{|x|}|x-3|<\frac{1}{3}\cdot\frac{1}{2}|x-3|<\epsilon~~\Rightarrow~~|x-3|<6\epsilon\]
which means you'd set \(\delta=\min(1,6\epsilon)\).

Answer 6

kitten