Precise limit definition Prove:
$$\displaystyle \lim_{x \to 2} \frac 1x = \frac 12.$$
We need to find a $\delta$ in terms of $\epsilon$.
Here is what I did so far:
\begin{align}
\left|\frac 1x-\frac 12 \right| &< \epsilon \\
-\epsilon < \frac 1x-\frac 12 &< \epsilon \\
-\epsilon + \frac 12< \frac 1x&< \epsilon + \frac 12 \\
\frac 2{1+2\epsilon} < x &< \frac 1{1-2\epsilon}
\end{align}
I am having trouble with this last step. Did I do this correctly?
 A: For your argument to work, you'll need $\epsilon < 1/2$, to ensure that $1 - 2\epsilon > 0$ and the last inequality makes sense. After that last step, you may set $\delta$ to be the distance from $2$ to the nearer endpoint of $(2/(1 + 2\epsilon), 1/(1 - 2\epsilon))$, i.e., $$\delta = \min\{2 - 2/(1 + 2\epsilon), 1/(1 - 2\epsilon) - 2\}.$$
For all $x$, if $|x - 2| < \delta$, then $x - 2 < \delta < 1/(1 - 2\epsilon) - 2$ and $2 - x < \delta <  2 - 2/(1 + 2\epsilon)$. This reduces to $2/(1 + 2\epsilon) < x < 1/(1 - 2\epsilon)$, which implies $|\frac{1}{x} - \frac{1}{2}| < \epsilon$ by your previous analysis.
A: pick an $\epsilon$ positive and smaller than $1/2.$ let $x_l, x_r$ be defined by $\dfrac{1}{x_l} = \dfrac{1}{2} + \epsilon, \dfrac{1}{x_r} = \dfrac{1}{2} - \epsilon.$ computing $x_r- 2 = \dfrac{2}{1-2\epsilon} - 2=\dfrac{4\epsilon}{1-2\epsilon}$  and  $2 - x_l = 2-\dfrac{2}{1+2\epsilon} =\dfrac{4\epsilon}{1+2\epsilon}$ 
choose $\delta = 4\epsilon$  so that $min \left\{2-x_l, 2-x_r   \right\} = 2 - x_l < \delta.$ use the fact that $\dfrac{1}{x}$ is decreasing for $x > 0$ to conclude $$\lvert \dfrac{1}{x} - \dfrac{1}{2} \rvert < \epsilon \text{ for all } 2 -\delta < x < 2 + \delta.$$
A: Did I do this correctly? Yes, you are correct.
(1) When you turn the expressions in both sides upside down, you change the signs (< or >) from left to right. 
(2) When you times (-1) to both sides, you also change the signs (< or >) from left to right. 
