Is this proof of $\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$ correct?

Question

I want to show that $\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$. Is this proof correct?

Let $\epsilon > 0$ be given and choose $\delta = \epsilon$. For $0 < |x-2| < \delta$, we have

$|\frac{1}{x} - \frac{1}{2}| = |\frac{2-x}{2x}|< \frac{|x-2|}{|x|} < |x-2| < \delta = \epsilon$

QED.

Answer 1

No, as explained by others. (Try $\varepsilon = \delta = 1.9$, $x = 0.1$.) You know what the derivative of $\frac{1}{x}$ is around $x = 2$, so you should know the ratio of $\delta$ to $\varepsilon$ in a neighborhood of $x = 2$.

Answer 2

For the following inequality $$ \frac{|x-2|}{|x|}<|x-2| $$ to hold, you need $$ |x|>1 $$ $\delta<1$. I would reccomend taking $\delta=\min\{ \epsilon,1\}$ to rectify this.

Answer 3

No, it's not correct. You used $1/|x|<1$, but it does not follow from your assumptions.

You need some bound on $1/x$. To do so, you can for example choose $\delta<1$, so that $x>1$. That is, you need to make $\delta$ small enough to avoid the trouble near $x=0$. Can you finish the proof with this modification?