1. ## Epsylons and deltas

let f(x) = 1/x for x>0. If E is any positive quantity (i.e. E >0), find a positive number S (i.e. S>0) which is such that :
if 0<|x-3|<S, then |f(x) - 1/3| < E

Note: E - epsylon, and S - delta

Am not sure of how to start. Any help would be appreciated.

2. Solve $\left|f(x) - \frac{1}{3}\right| < \epsilon$ for $|x - 3|$.

$\left|f(x) - \frac{1}{3}\right| < \epsilon$

$\left|\frac{1}{x} - \frac{1}{3}\right| < \epsilon$

$\left|\frac{3}{3x} - \frac{x}{3x}\right| < \epsilon$

$\left|\frac{3 - x}{3x}\right| < \epsilon$

$\frac{|3 - x|}{|3x|} < \epsilon$

$\frac{1}{3|x|}|x-3| < \epsilon$

$\frac{1}{|x|}|x - 3| < 3\epsilon$.

Define M such that $\frac{1}{|x|} \leq M$, then

$|x - 3| < \frac{3\epsilon}{M}$.

So we would be able to choose $\delta = \frac{3\epsilon}{M}$.

There is no $M$ such that $\frac{1}{|x|} < M$ for all $x$, but we can restrict the value of $x$.

If we restrict $|x - 3| < 1$ (i.e. so that $\delta = 1$)

then $-1 < x - 3 < 1$

$2 < x < 4$

$\frac{1}{4} < \frac{1}{x} < \frac{1}{2}$.

Thus $\frac{1}{|x|} < \frac{1}{2}$.

So set $M = \frac{1}{2}$ so that we can define $\delta$ to be

$\delta = \min\left\{1, \frac{3\epsilon}{\frac{1}{2}}\right\}$

$\delta = \min\{1, 6\epsilon\}$.

Now all that's left is to write the proof.

Proof: Let $\epsilon > 0$ and define $\delta = \min\{1, 6\epsilon\}$. Then if $0 < |x - 3| < \delta$, we have

$\left|\frac{1}{x} - \frac{1}{3}\right| = \left|\frac{3}{3x} - \frac{x}{3x}\right|$

$= \left|\frac{3 - x}{3x}\right|$

$= \frac{|x - 3|}{3|x|}$

$< \frac{1}{2}\left(\frac{|x - 3|}{3}\right)$ since $\frac{1}{|x|} < \frac{1}{2}$

$< \frac{1}{2}\left(\frac{6\epsilon}{3}\right)$ since $|x - 3| < \delta$ and $\delta < 6\epsilon$

$= \epsilon$.

Thus $0 < |x - 3| < \delta \implies \left|f(x) - \frac{1}{3}\right| < \epsilon$.