Prove $\lim_{x \to 2} \frac{1}{x} = \frac{1}{2}$ using delta epsilon definition.
So I did $|x-2|<\delta$ and $|1/x -1/2|<\varepsilon$.
The second inequality can be written as $|(2-x)/2x|<\varepsilon$ and this can be written as $|(x-2)/2x|<\varepsilon$ but now there is a $2x$ in the denominator.
Because I could write $|x-2|<\varepsilon 2x$ but then delta depends on x if I choose $\delta= \varepsilon 2x$. I could take a nearby point and say $|x-2|<1$ so $1<x<3$ then delta would be $6\varepsilon$ this also doesnt seem right.
Best Answer
We need to show that given any $\varepsilon>0$ that there exists a number $\delta>0$ such that whenever $0<|x-2|<\delta$, $\left|\frac1x-\frac12\right|<\varepsilon$.
Since we are interested in the behavior of $\frac1x$ when $x$ is "close" to $2$, let's limit the interval over which we examine such behavior. So, let's agree to look in the interval $1<x<3$ or $|x-2|<1$. While this is arbitrary, it is fit for purpose here.
Now that we've agreed to restrict $x$ to the interval $(1,3)$, we can write
$$\left|\frac1x-\frac12\right|=\frac{|x-2|}{|2x|}<\frac12 |x-2|\tag1$$
The right-hand side of $(1)$ is less than and given $\varepsilon>0$ when $|x-2|<2\varepsilon$.
Hence, if we take $\delta>0$ as the smaller of $1$ and $2\varepsilon$, then we can assert that for any $\varepsilon>0$, there exists a $\delta=\min(1,2\varepsilon)$ such that
$$\left|\frac1x-\frac12\right|<\varepsilon$$
whenever $0<|x-2|<\delta$.
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