Limit proof, finding upper bounds I need to prove that $\frac{1}{x-3} \to 1$ as $x \to 4$
So I did $\left | \frac{1}{x-3} - 1 \right | = \left | \frac{4 - x}{x-3}  \right | = \left | \frac{x-4}{x-3} \right |  = \frac{|x-4|}{|x-3|} <  K|x-4|$. So I need to pick $| x - 4| < \delta = \frac{\epsilon}{K}$
Now the problem is that I can't bound my $\frac{1}{|x-3|}$

Let $\epsilon>0$. Suppose $|x-4|<1/2$. Then $-1/2<x-4<1/2$ which implies that  $1/2<x-3<3/2$. Hence,  $|x-3|\ge x-3>1/2$, that is, if $x\ne 3$ then $\frac{1}{|x-3|}<2$. Define $\delta=\min\left\{\frac{1}{2},\frac{\epsilon}{2}\right\}$. Let $0<|x-4|<\delta$. Then certainly $x\ne 3$. Hence,
$\left| \frac{1}{x-3}-1\right|=\left| \frac{x-4}{x-3}\right|<|x-4|\cdot 2 < \frac{\epsilon}{2}\cdot 2=\epsilon$.
The result follows.
