2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ He multiplies numerator and denominator by $2x$. $\endgroup$ – Gerry Myerson Jun 9 '12 at 6:33
  • $\begingroup$ But how does he know to do this? I would love to know how he knew to multiply both by 2x, cheers :) $\endgroup$ – Joseph Jun 9 '12 at 6:34
  • $\begingroup$ It's standard: in order to add fractions, you have to multiply the numerator and denominator of each term by a number (not necessarily the same for each term) in such a way that they all end up with the same denominator. When we have reciprocals of objects that share no common factor (like a formal variable and the number $2$), we simply multiply by their product: $$\frac{1}{2}-\frac{1}{x}=\frac{x}{2x}-\frac{2}{2x}=\frac{x-2}{2x}.$$ $\endgroup$ – anon Jun 9 '12 at 6:37
  • $\begingroup$ You've got a fraction inside of a fraction. Nobody likes those. What to do? Well, if you multiply the numerator by 2, one of those fractions in the numerator goes away. Of course, you can't just multiply the numerator by 2, that would change the value of the fraction; you have to multiply both numerator and denominator by 2. But then there's still the other fraction in the numerator. I can make that go away by multiplying by $x$ --- again, in both numerator and denominator. And as long as I'm going to multiply by 2 and by $x$, might as well do both at once --- just multiply by $2x$. $\endgroup$ – Gerry Myerson Jun 9 '12 at 12:39
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Cheers, this cleared it up. Darn Basic fraction skills. :) $\endgroup$ – Joseph Jun 9 '12 at 6:47
0
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ Thanks, all the answers here helped, cheers. $\endgroup$ – Joseph Jun 9 '12 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.