Where am I going wrong when trying to prove $\lim _{ x\rightarrow \infty }{ \frac { 1-2x }{ 3-4x } =\frac { 1 }{ 2 } } $ $$\lim _{ x\rightarrow \infty  }{ \frac { 1-2x }{ 3-4x } =\frac { 1 }{ 2 }  } $$
Proof: Let $\epsilon > 0$
Then, $$\left| \frac { 1-2x }{ 3-4x } -\frac { 1 }{ 2 }  \right| <\epsilon$$
$$\Longleftrightarrow \left| \frac { 2-4x }{ 6-8x } -\frac { 3-4x }{ 6-8x }  \right| <\epsilon $$
$$\Longleftrightarrow \left| \frac { -1 }{ 6-8x }  \right| <\epsilon $$
$$\Longleftrightarrow \frac { 1 }{ \left| 6-8x \right|  } <\epsilon $$
Now we will calculate the lower bound on $x$ to force: $\Longleftrightarrow \left| \frac { 2-4x }{ 6-8x } -\frac { 3-4x }{ 6-8x }  \right| <\epsilon $ to hold true
Without loss of generality, assume that $x>\frac { 3 }{ 4 } $
Then, we get: $$\frac { 1 }{ 6-8x } >\epsilon$$
$$\Longrightarrow 1>\epsilon (6-8x)$$
$$\Longrightarrow 1>\epsilon 6-\epsilon 8x$$
$$\Longrightarrow \epsilon 8x>\epsilon 6-1$$
$$\Longrightarrow x>\frac { 3 }{ 4 } -\frac { 1 }{ 8 \epsilon} $$
Therefore, since $x>\frac { 3 }{ 4 } -\frac { 1 }{ 8 \epsilon} $, we can conclude that : $$\left| \frac { 1-2x }{ 3-4x } -\frac { 1 }{ 2 }  \right| <\epsilon \Longleftrightarrow x>\frac { 3 }{ 4 } -\frac { 1 }{ 8 \epsilon} $$
I flipped the sign after the "without loss of generality..." line because I figured that $$x>\frac { 3 }{ 4 } $$ would make $\frac { 1 }{ 6-8x } >\epsilon$ negative. I am unsure of my steps starting with the line where I put "without loss of generality..." so I need clarification as how I should proceed from there because I feel like what I did here is wrong. 
 A: There were actually at least two errors after you assumed $x > \frac 34$.
You were correct that changing $\frac{1}{\lvert 6-8x \rvert}$
to $\frac{1}{6-8x}$ is a sign change.
It is also true that you can change the signs of both sides of an inequality
if you reverse the direction of the inequality.
But that's if you change both sides.
What you actually did was to change the sign on the left side but not
on the right side. So your next inequality was not equivalent to
$$\frac{1}{\lvert 6-8x \rvert} < \epsilon.$$
An inequality that is equivalent, and looks a lot like what you
wrote (except for one negative sign), is
$$\frac{1}{6-8x} > -\epsilon.$$
Next, you multiplied both sides by $6 - 8x$. But remember, $6 - 8x$ is negative,
and multiplying both sides of an inequality by a negative number flips the
direction of the inequality (again), so you should have
$$1 < -(6-8x)\epsilon$$
or more simply,
$$1 < (8x - 6)\epsilon,$$
which is what you would have gotten much more easily if you had simply substituted $8x - 6$ for $\lvert 6-8x \rvert$ in the first place.
