Prove $\lim_{x \to 2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}=\frac{-1}{4}$ using $\epsilon-\delta $ condition

Solution 1:

From what you have deduced, you can easily check that:

$ \displaystyle\left|\frac{\displaystyle\frac{1}{x}-\frac{1}{2}}{x-2}+\frac{1}{4}\right|<\epsilon\Leftrightarrow -\epsilon-\frac{1}{4}<-\frac{1}{2x}<\epsilon-\frac{1}{4}\Leftrightarrow \frac{1}{2\epsilon+\displaystyle\frac{1}{2}}<x<\frac{1}{-2\epsilon+\displaystyle\frac{1}{2}}\Leftrightarrow \frac{2}{4\epsilon+1}<x<\frac{2}{-4\epsilon+1}$

So, you now have that: $\delta \leq \min{\displaystyle \{\frac{2}{-4\epsilon+1},\frac{2}{4\epsilon+1}\}}$. Since $\epsilon>0,$ set $\delta \displaystyle\leq \frac{2}{-4\epsilon+1}$.

But because we want $\delta > 0$, we must have

$\displaystyle\frac{2}{-4\epsilon+1}>0\Leftrightarrow \epsilon <\frac{1}{4}$

Solution 2:

Take $\delta = \min (1, 4 \epsilon)$. Then for $\vert x-2\vert \lt \delta$, you have (using reverse triangle inequality)

$$2 - \vert x \vert\lt \vert x -2\vert \le 1$$ and therefore $\vert x \vert \gt 1$. Which leads to

$$\frac{\vert x-2 \vert}{4\vert x \vert} \le \frac{\vert x-2 \vert}{4} \lt \frac{\delta}{4} \le \epsilon $$