<s>[INST] Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$ I was working on a calculus problem:
Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$
I did some algebra on the consequent obtaining:
$$
\frac{|x-3|}{|3x|} < 10^{-4} 
$$
Then I noted that if $0< |x-3| < 1$ for $\delta = 1$. We have:
$$
\frac{|x-3|}{|3x|} < \frac{1}{6}  
$$
So it's  sufficient to have $\delta = 6\cdot 10^{-4} $, 
$$
\frac{|x-3|}{|3x|} < \frac{|x-3|}{6} < \frac{1}{6} \cdot 6\cdot 10^{-4} 
$$
Basically this follows because we have a lower bound on the denominator. Is this reasoning good, or is there something I'm not accounting for? 
 [/INST] Note that
$0<|x-3|<\delta\iff x\in(3-\delta, 3+\delta)\iff 3-\delta < x < 3+\delta$.  
Therefore, you can write $x = 3 + \rho$, such that $|\rho| < \delta$. So you have that
$\bigg\lvert \frac{1}{x}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{1}{3 + \rho}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{3- 3-\rho}{3(3+\rho)}\bigg\lvert=\bigg\lvert\frac{\rho}{3(3+\rho)}\bigg\lvert$.
If you take $\delta$ such that $\delta\leq10^{-4}$, then $|\rho| <\delta\Rightarrow \bigg\lvert \frac{1}{x}-\frac{1}{3}\bigg\lvert =\bigg\lvert\frac{\rho}{3(3+\rho)}\bigg\lvert < \frac{\delta}{3\cdot2}<\frac{10^{-4}}{6}<10^{-4}$.
</s>