Prove that $\lim_{x\to2}\frac1x=\frac 12$

Question

So far:

$$\begin{align}f(x) &= \frac1x\\ L &= \frac12\\ a &= 2 \end{align} $$

$0<|x-2|<\delta$ and $\left|\frac1x - \frac12\right|<\epsilon$

I have been trying to manipulate the $|f(x) - L|$ function to be in the form of the $|x-a|$ function, but have not been able to figure it out. I cannot figure out how to make further progress. Thanks!

Answer 1

Let $\epsilon>0$ be a real number.

Then let $\delta := \min\{1,2\epsilon\}$.

If $x\in\mathbb R$ fulfils $|x-2|<\delta$, then in particular $|x|>1$. Thus,

$$\left|\frac 1x-\frac 12 \right| = \left|\frac{x-2}{2x} \right| < \frac{\delta}{2|x|}\leq \frac{\epsilon}{|x|} < \epsilon$$

Thus, by definition of convergence, $\lim_{x\rightarrow 2} \frac 1x=\frac 12$.