calculus - Proving that $limlimits_{xto 4}frac{2x-5}{6-3x} = -frac{1}{2} $

Proving that $$\lim_{x\to 4}\frac{2x-5}{6-3x} = -\frac{1}{2} $$

Now of course when you plug in $4$ you get $-\frac{1}{2}$, but I need to prove it using:

$\forall \varepsilon \gt 0,\;\exists\delta \gt 0\;\forall x \in D $ (where $D$ is the domain of the function)

$$0\lt |x-x_0|\lt \delta \implies |f(x)-L| \lt \varepsilon $$

Where $L$ is the limit we "guessed" and $x_0$ is where $x$ approaches)

So far I have:

$$\begin{align}
0\lt |x-4|\lt \delta &\implies \left|f(x)-\left(-\frac{1}{2}\right)\right| \lt \varepsilon\\
\iff 0\lt |x-4|\lt \delta &\implies \left|\frac{2x-5}{6-3x}+\left(\frac{1}{2}\right)\right| \lt \varepsilon\\
\iff 0\lt |x-4|\lt \delta &\implies \left|\frac{4-x}{6(x-2)}\right| \lt \varepsilon

\end{align}$$

This is where I'm stuck. What exactly do I need to prove after this last line?

Answer

Given any $\varepsilon > 0$, let $\delta = \min\{1, \varepsilon/M\} > 0$, where $M$ is some fixed magical positive constant. While doing this proof, we will eventually figure out what $M$ should be, then go back and erase $M$ and replace it with this constant. Now if $0 < |x - 4| < \delta$, then observe that:
\begin{align*}
\left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right|
&= \left| \frac{4 - x}{6(x - 2)} \right| \\
&= \frac{1}{6|x - 2|}|x - 4| \\
&< \frac{1}{6|x - 2|} \cdot \frac{\varepsilon}{M} &\text{since } |x - 4| < \delta \leq \frac{\varepsilon}{M} \\

\end{align*}
Now we take a break from our proof and try to figure out a way to bound $\frac{1}{6|x - 2|}$ by a constant $M$ given that $x$ is at most $1$ away from $4$, implying that $|x - 2|$ won't be too close to zero. Indeed, if $|x - 4| < 1$, then observe that:
\begin{align*}
|x - 2|
&= |(x - 4) - (-2)| \\
&\geq ||x - 4| - \left| -2 \right|| &\text{by the reverse triangle inequality} \\
&= 2 - |x - 4| &\text{since } |x - 4| < 1 \leq 2 \\
&> 2 - 1 &\text{since } |x - 4| < 1 \\
&= 1
\end{align*}

so that:
$$
\frac{1}{6|x - 2|} < \frac{1}{6}
$$
Thus, by taking $M = \frac{1}{6}$, we can finish up our original proof:

---

Given any $\varepsilon > 0$, let $\delta = \min\{1, 6\varepsilon\} > 0$. Then if $0 < |x - 4| < \delta$, observe that:
\begin{align*}

\left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right|
&= \left| \frac{4 - x}{6(x - 2)} \right| \\
&= \frac{1}{6|x - 2|}|x - 4| \\
&< \frac{1}{6|x - 2|} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 6\varepsilon \\
&< \frac{1}{6} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 1 \\
&= \varepsilon
\end{align*}
as desired. $~\blacksquare$