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- Thread starter Mr Davis 97
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Mark44

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No, you get a different expression whose value is the same.This is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression.

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?

Without an example, I'm not sure what you're asking about. Here's an example that might answer your question.

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).

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

Without an example, I'm not sure what you're asking about. Here's an example that might answer your question.

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.

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Mark44

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Sometimes people will expicitly write any restrictions to the domain. For example,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.

##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.

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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?

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Mark44

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No, that's not the only way to evaluate3 this limit. You can also factor both numerator and denominator like so: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}}##.

## \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 thatMr Davis 97 said:However, this changes the domain of the original function such that x = 0 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?

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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}}##?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 thatatx= 0the original functionis 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.

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Mark44

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Of course they are equal,But ##\displaystyle \frac{2x - 1}{x + 1} \neq \frac{2 - \frac{1}{x}}{1 + \frac{1}{x}}##.

I don't understand why you're so hung up on this.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}}##?

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HallsofIvy

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I want to expand on what Mark44 said. You do NOT get "the same expression", you get another expression with the same numeric valueThis is a simple question. I know that for any number or expression we can multiply it by 1 and maintain the same expression.i

Yes, because of your misquoting of that "rule"- you should have included "if x n is not 0" right at the start.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?

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So if we divide the x's out again to get the original expression, that tacitly adds 0 back to our solution set?

As you can't divide by zero it is not that simple. But you can remove the singularity using L'Hôpital's rule.

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