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.
Best Answer
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.