# What does it mean to multiply by x / x?

This is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression. However, x / x also equals 1. But if we multiply it by another expression, the expression changes such that we can no longer use 0. So if the domain of an expression originally contains 0 and we multiply it by x / x, is that illegal?

Mark44
Mentor
This is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression.
No, you get a different expression whose value is the same.
Mr Davis 97 said:
However, x / x also equals 1. But if we multiply it by another expression, the expression changes such that we can no longer use 0. So if the domain of an expression originally contains 0 and we multiply it by x / x, is that illegal?

x(x - 1) = 0 -------------- Solution set {0, 1}
If we multiply the left side by x/x, we get
##\frac{x^2(x - 1)}{x} = 0## -------------- The numerator is zero if x = 0 or x = 1, but we can't divide by 0, so the solution set is now {1}.

By multiplying by x/x, we are tacitly assuming that ##x \ne 0##, so we lose a value in the solution set.

If we graph y = x(x - 1) and ##y = \frac{x^2(x - 1)}{x}##, the two graphs look exactly the same, except that the graph of the latter equation has a "hole" at (0, 0).

symbolipoint
No, you get a different expression whose value is the same.

x(x - 1) = 0 -------------- Solution set {0, 1}
If we multiply the left side by x/x, we get
##\frac{x^2(x - 1)}{x} = 0## -------------- The numerator is zero if x = 0 or x = 1, but we can't divide by 0, so the solution set is now {1}.

By multiplying by x/x, we are tacitly assuming that ##x \ne 0##, so we lose a value in the solution set.

If we graph y = x(x - 1) and ##y = \frac{x^2(x - 1)}{x}##, the two graphs look exactly the same, except that the graph of the latter equation has a "hole" at (0, 0).

So if we divide the x's out again to get the original expression, that tacitly adds 0 back to our solution set? The only thing that confuses me is that the addition or subtraction of the domain element seems to have to be inferred by us rather than explicitly stated by the mathematics.

Mark44
Mentor
So if we divide the x's out again to get the original expression, that tacitly adds 0 back to our solution set? The only thing that confuses me is that the addition or subtraction of the domain element seems to have to be inferred by us rather than explicitly stated by the mathematics.
Sometimes people will expicitly write any restrictions to the domain. For example,
##y = \frac{x^2(x - 1)}{x}, x \ne 0##
##\Leftrightarrow y = x(x - 1), x \ne 0##
The two equations above are equivalent, meaning that their solution sets (and graphs) are identical.

For examples as simple as this one, some writers will omit the explicit domain restrictions, believing them to be pretty much obvious.

Sometimes people will expicitly write any restrictions to the domain. For example,
##y = \frac{x^2(x - 1)}{x}, x \ne 0##
##\Leftrightarrow y = x(x - 1), x \ne 0##
The two equations above are equivalent, meaning that their solution sets (and graphs) are identical.

For examples as simple as this one, some writers will omit the explicit domain restrictions, believing them to be pretty much obvious.

Okay, that makes sense. But what about when we are solving limits at infinity? For example, given, ##\displaystyle \lim_{x\rightarrow \infty } \frac{2x - 1}{x + 1}##, the only way to solve is to multiply the fraction by ##\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}##. However, this changes the domain of the original function such that x = 0 is no longer defined. So why is this a rigorous way to solve limits at infinity if we're changing the domain of the function?

Mark44
Mentor
Okay, that makes sense. But what about when we are solving limits at infinity? For example, given, ##\displaystyle \lim_{x\rightarrow \infty } \frac{2x - 1}{x + 1}##, the only way to solve is to multiply the fraction by ##\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}##.
No, that's not the only way to evaluate3 this limit. You can also factor both numerator and denominator like so:
## \lim_{x\to \infty } \frac{2x - 1}{x + 1} = \lim_{x\to \infty } \frac{x(2 - 1/x)}{x(1 + 1/x)} = \lim_{x\to \infty } \frac x x \cdot \lim_{x\to \infty }\frac{x(2 - 1/x)}{x(1 + 1/x)}##.
The first limit is 1 and the second limit it 2. In the first limit, x/x is always 1 for any finite value of x (other than 0), so as x increases without bound, the value of the fraction is always 1.
Mr Davis 97 said:
However, this changes the domain of the original function such that x = 0 is no longer defined.
Corrected the above to "such that at x= 0 the original function is no longer defined."
It doesn't matter. You're taking the limit as ##x \to \infty##, so the fact that the rational expression isn't defined at x = 0 is immaterial.
Mr Davis 97 said:
So why is this a rigorous way to solve limits at infinity if we're changing the domain of the function?

No, that's not the only way to evaluate3 this limit. You can also factor both numerator and denominator like so:
## \lim_{x\to \infty } \frac{2x - 1}{x + 1} = \lim_{x\to \infty } \frac{x(2 - 1/x)}{x(1 + 1/x)} = \lim_{x\to \infty } \frac x x \cdot \lim_{x\to \infty }\frac{x(2 - 1/x)}{x(1 + 1/x)}##.
The first limit is 1 and the second limit it 2. In the first limit, x/x is always 1 for any finite value of x (other than 0), so as x increases without bound, the value of the fraction is always 1.
Corrected the above to "such that at x= 0 the original function is no longer defined."
It doesn't matter. You're taking the limit as ##x \to \infty##, so the fact that the rational expression isn't defined at x = 0 is immaterial.
But ##\displaystyle \frac{2x - 1}{x + 1} \neq \frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}##. So how can I be sure that ##\displaystyle \lim_{x\rightarrow \infty }\frac{2x - 1}{x + 1} = \lim_{x\rightarrow \infty }\frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}##?

Mark44
Mentor
But ##\displaystyle \frac{2x - 1}{x + 1} \neq \frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}##.
Of course they are equal, except at x = 0 and x = -1. Again, the limit is as ##x \to \infty##, so for any x > 0, the two expressions are identical.
Mr Davis 97 said:
So how can I be sure that ##\displaystyle \lim_{x\rightarrow \infty }\frac{2x - 1}{x + 1} = \lim_{x\rightarrow \infty }\frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}##?
I don't understand why you're so hung up on this.

It's easier to do polynomial division. ## \frac{2x -1}{ x + 1} \rightarrow 2 + \frac{-3}{x + 1} ##. Taking ## \frac{\lim}{x\rightarrow \infty } \rightarrow 2 ##. What your worried about is the equivalence of fractions, but if you can show the steps in light of the Properties of Equations (e.g., Division Property of Equations) then you have nothing to worry about.

HallsofIvy