Show the limit holds true. I have to show that: lim$_{x\rightarrow 2} \displaystyle\frac{1}{x} = \frac{1}{2}$

So far I have: 

$|f(x)-L|&lt;\epsilon \leftrightarrow |\displaystyle\frac{1}{x}-\frac{1}{2}|&lt;\epsilon \leftrightarrow |\displaystyle\frac{x-2}{2x}|&lt; \epsilon \leftrightarrow \frac{1}{2x} \cdot |x-2| &lt; \epsilon \leftrightarrow |x-2| &lt; \frac{2x}{\epsilon}$. So $\delta=\displaystyle\frac{2x}{\epsilon}$. 

My question is: Is it okay for $\delta$ to depend on $x$?
Remember that by definition

$$\lim_{x\rightarrow a} f(x) = L \iff \forall \varepsilon &gt;0\, \exists \delta &gt; 0: 0&lt;\vert x-a\vert &lt;\delta \implies \vert f(x)-L\vert &lt;\varepsilon$$

To find the optimal $\delta$ you can proceed as follow

$$\left|\frac1x-\frac12 \right|&lt;\epsilon \iff -\epsilon&lt;\frac1x-\frac12 &lt;\epsilon\iff -\epsilon+\frac12&lt;\frac1x&lt;\epsilon+\frac12$$

$$\iff \frac{1-2\epsilon}2&lt;\frac1x&lt;\frac{1+2\epsilon}2\iff \frac2{1+2\epsilon}&lt;x&lt;\frac2{1-2\epsilon}$$

$$\iff \frac2{1+2\epsilon}-2&lt;x-2&lt;\frac2{1-2\epsilon}-2$$
$$\iff \frac{-4\epsilon}{1+2\epsilon}&lt;x-2&lt;\frac{4\epsilon}{1-2\epsilon}$$

thus

$$|x-2|&lt;min\{\frac{4\epsilon}{1-2\epsilon},\frac{4\epsilon}{1+2\epsilon} \}=\frac{4\epsilon}{1+2\epsilon}=\delta$$
No, it is not okay if your $\delta$ depends on $x$. It can depend on $x_0$, in your case $x_0 = \dfrac{1}{2}$.

But why do you need a $\delta$ or do you want to proof continuity?