How does he get a perfect swap numerator and denominator. I'm going through a exercise, in which all the answers are given, but the tutor makes a step and I can't follow at all. A massive jump with no explanation.
Here is the question:
$\lim_{x \to 2} \frac{\frac{1}{2}-\frac{1}{x}}{x-2}$
He then simplifies:
$ \frac{x-2}{2x(x-2)}$
He said he multiplied the entire equation by 2x
How does he know 2x swaps the denominator upto the numerator 
Before he gave the simplification I spent perhaps 20 minutes trying to figure something out, and failed, and then he just jumps this massive step
He simplifies because we are finding limits.
Thanks
Joseph G.
 A: If he said that he multiplied the expression by $2x$, he misspoke. He multiplied it by $\frac{2x}{2x}$. Note that $\frac{2x}{2x}=1$, so that it’s entirely permissible to multiply by it, while multiplying by $2x$ would change the value.
He took a small shortcut. I’ll do it the long way first, putting the numerator over a common denominator and simplifying the resulting three-story fraction:
$$\begin{align*}
\frac{\frac12-\frac1x}{x-2}&=\frac{\frac12\cdot\frac{x}x-\frac1x\cdot\frac22}{x-2}\\\\
&=\frac{\frac{x}{2x}-\frac2{2x}}{x-2}\\\\
&=\frac{\frac{x-2}{2x}}{x-2}\\\\
&=\frac{\frac{x-2}{2x}}{x-2}\cdot\frac{2x}{2x}\tag{1}\\\\
&=\frac{x-2}{2x(x-2)}\;.
\end{align*}$$
The tutor merely observed that when the numerator is put over a common denominator, that denominator will be $2x$, and avoided the first few steps of my calculation by essentially going directly to the step marked $(1)$:
$$\begin{align*}
\frac{\frac12-\frac1x}{x-2}&=\frac{\frac12-\frac1x}{x-2}\cdot\frac{2x}{2x}\\\\
&=\frac{\left(\frac{x}{2x}-\frac2{2x}\right)2x}{2x(x-2)}\\\\
&=\frac{x-2}{2x(x-2)}\;.
\end{align*}$$
There’s no swapping of the denominator into the numerator: it just happens that when the numerator is simplified, the result is a fraction whose numerator is the same as the original denominator.
A: Note that
\begin{align*}
\frac{\frac{1}{2}-\frac{1}{x}}{x-2} &= \frac{\frac{x-2}{2x}}{x-2} \\\ &= \frac{2x \cdot \frac{x-2}{2x}}{2x \cdot (x-2)} \qquad \Bigl[\small\text{Multiplying numerator and denominator by} \ 2x \Bigr] \\\ &= \frac{(x-2)}{2x \cdot (x-2)} \\\ &= \frac{1}{2x}
\end{align*}
