Difficulties with estimation of epsilon-delta limit proof I have to proof $\lim_{x\to 5}\frac{4x-9}{3x-16}=-11$. I have hard time to evaluate $\frac{1}{|3x-16|}$.
So I start that let $|x-5|<$. I need to show that $|\frac{4x-9}{3x-16}-(-11)|<ε$.
$|\frac{4x-9}{3x-16}-(-11)|$ = $|\frac{37x-185}{3x-16}|$ = $\frac{5|x-5|}{|3x-16|}$ < $ε$.
I choose $:=1$, then $|x-5|<1$, where I get $-1 < x-5 < 1$, then multiply with 3 and subtract -1, so $-4 < 3x-16 < 2$. If I think about it as $\frac{-1}{4} > \frac{1}{3x-16} > \frac{1}{2}$ contradiction.
Is it wrong to continue the way that $ |\frac{1}{3x-16}| < \frac{1}{2}$ ? So I get that $\frac{5|x-5|}{|3x-16|}$< $5 *  *  \frac{1}{2}$ <= $ε $ if $ = \frac{2}{5}*ε$.
Finaly I got that := min {1, $\frac{2}{5}*ε$}
I would need some help with $\frac{1}{|3x-16|}$.
 A: A systematic advice in such a situation is to "translate the variable to $0$", i.e. here: let
$$x=5+h,$$
so that $x\to5\iff h\to0.$ If $x\ne\frac{16}3,$
$$\frac{4x-9}{3x-16}+11=\frac{37h}{-1+3h}.$$
If $h<\frac13$ then
$$\begin{align}\left|\frac{37h}{-1+3h}\right|<\varepsilon&\iff37|h|<\varepsilon(1-3h)\\&\Longleftarrow37|h|<\varepsilon(1-3|h|)\\&\iff|h|<\frac\varepsilon{37+3\varepsilon}.
\end{align}$$
A convenient $\delta$ is therefore $\min\left(\frac13,\frac\varepsilon{37+3\varepsilon}\right).$
Note that José's $\delta=\min\left(\frac\varepsilon{74},\frac16\right)$  is smaller, hence also convenient.
A: If $|x-5|<\frac16$, then $5-\frac16<x<5+\frac16$, and therefore $-\frac32<3x-16<-\frac12$. So, $|3x-16|>\frac12$.
So, if you take $\delta\leqslant\frac16$ and if $|x-5|<\delta$, you have$$\left|\frac{4x-9}{3x-16}-(-11)\right|=\frac{37|x-5|}{|3x-16|}<74|x-5|.$$So, you can take $\delta=\min\left\{\frac\varepsilon{74},\frac16\right\}$, in order that$$\left|\frac{4x-9}{3x-16}-(-11)\right|<\varepsilon.$$
In your answer, you had $-4<3x-16<2$ and you deduced that $-\frac14>\frac1{3x-16}>\frac12$, which is wrong. You have indeed $a<b<c\implies\frac1a>\frac1b>\frac1c$, but only when $a>0$ or $c<0$.
