Proving the limit $\lim_{x\to 1} \frac{100}{x} = 100$ using epsilon-delta definition $$\lim_{x\to 1} \frac{100}{x} = 100$$
I'm trying to proving the limit using epsilon-delta definition and here's what I've come up with so 
My attempt:
$\forall ε>0, \exists δ>0$, such that if for x, $0<|x-1|<δ$ then $|\frac{100}{x} - 100| < ε$
$$100*|\frac{1}{x} - 1| < ε$$
$$100*\frac{1}{|x|} * |x-1| < ε$$
δ : $$100*\frac{1}{|x|} * |x-1| < ε$$
$$|x-1| < δ$$, choose δ = 1/2
then,
$$|x-1| < \frac{1}{2} \Rightarrow 0 < x < \frac{3}{2}$$
From here I'm not too sure how to proceed on. I was wondering if anyone can help me on what to do next.

You want $$\left|\frac{100}{x} - 100\right| < \varepsilon$$
whenever $\left|x-1\right| < \delta$.
Your task is to find a $\delta$ that makes this true for a given $\varepsilon.$
Let $\varepsilon > 0$, we have
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\left|\frac{1}{x} - 1\right| =
100\cdot\left|\frac{1-x}{x}\right| =
100\cdot\frac{\left|1-x\right|}{\left|x\right|}.
\end{align}
If we choose $\delta\le\frac 1 2$ we have $\left|x-1\right|<\frac 1 2$ which says
$$-\frac 1 2 < x - 1 < \frac 1 2$$
or equivalently
$$ \frac 1 2 < x < \frac 3 2.$$
In particular $\left|x\right|=x > \frac 1 2$, so we continue with
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\frac{\left|1-x\right|}{\left|x\right|} =
100\cdot\frac{\left|1-x\right|}{\left|x\right|} \\&<
100\cdot\frac{\left|1-x\right|}{1/2} = 200 \cdot \left|1-x\right| < 200\cdot \delta.
\end{align}
For this to be less than or equal to $\varepsilon$ we need $200\cdot\delta\le\varepsilon$, which is equivalent to $\delta \le \frac{\varepsilon}{200}$.
So if we have both $\delta\le\frac 1 2$ and $\delta<\frac{\varepsilon}{200}$, which can be achieved by picking
$$ \delta = \min \left\{\frac 1 2, \frac{\varepsilon}{200}\right\},$$
we get
\begin{align}
\left|\frac{100}{x} - 100\right| &=
100\cdot\frac{\left|1-x\right|}{\left|x\right|} < 200\cdot \delta \le 200\cdot \frac{\varepsilon}{200} = \varepsilon.
\end{align}

You get this when you look at $$\frac{\left|1-x\right|}{\left|x\right|},$$ and realize that to get this fraction small you need $\left|1-x\right|$ to be small, which can be achievied directly by choosing $\delta$ small, but you also want $\left|x\right|$ to don't be small, since a small denominator gives a large fraction. So you need a way to make $\left|x\right|>c>0$ by making $\left|x-1\right|$ small. I chose to make $\left|x-1\right|<\frac 1 2$ which gave us $\left|x\right|>\frac 1 2$ so we can finish the proof.
