Justification for cancelling terms in limits?

[Mr Davis 97]

I am confused about the algebraic process of finding a limit. Let us take $\frac{x^2 -1}{x - 1}$. In trying to find $\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}$we do the following:

$\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}$

$\displaystyle\lim_{x\rightarrow 1}x+1$

$2$

But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that $\lim_{x\rightarrow 1}x+1$ leads to the correct answer to the original problem if $x + 2$ is different function, with a different domain, than $\frac{x^2 -1}{x - 1}$?

 

[Mark44]

I am confused about the algebraic process of finding a limit. Let us take $\frac{x^2 -1}{x - 1}$. In trying to find $\lim_{x\rightarrow 1}\frac{x^2 -1}{x - 1}$we do the following:

$\displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}$

$\displaystyle\lim_{x\rightarrow 1}x+1$

$2$

But what justification do we have for cancelling the (x - 1) terms?

Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction $\frac{x - 1}{x - 1}$ will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.

When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that $\lim_{x\rightarrow 1}x+1$ leads to the correct answer to the original problem if $x + 2$ is different function, with a different domain, than $\frac{x^2 -1}{x - 1}$?

The only difference between $y = x + 1$ and $y = \frac{x^2 - 1}{x - 1}$ is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.

 

[Qwertywerty]

How can we be absolutely certain that limx→1x+1\lim_{x\rightarrow 1}x+1 leads to the correct answer to the original problem if x+2x + 2 is different function, with a different domain, than x2−1x−1\frac{x^2 -1}{x - 1}?

I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .

For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .

If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10^-30 / 10^-30 is one .

 

[Mr Davis 97]

Since the limit is as x approaches 1, as long as x is not exactly equal to 1, the fraction $\frac{x - 1}{x - 1}$ will be equal to 1. We are dividing a nonzero number by itself. Passing to the limit doesn't change things.

The only difference between $y = x + 1$ and $y = \frac{x^2 - 1}{x - 1}$ is a single point of discontinuity in the graph of the second equation at (1, 2). Otherwise the graphs of the two equations are exactly the same.

I think you mean x + 1 . Okay - What is the only difference between the two graphs ? Their domain , as you have already said .

For all x ∈ R - { 1 } , they are essentially the same function . The latter function is not defined at x = 1 , and thus the only thing we can try to do at x = 1 , is find what value it would tend to .

If you have , say a variable z , such that f(z) = z/z . Then you can always cancel the z from the numerator and denominator , as long as z does not equal zero . 0 / 0 is an indeterminate form , but say , 2 / 2 , or even 10^-30 / 10^-30 is one .

I understand what both of you are saying, and it helps. But what I am asking is what guarantees that $\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}$ (for example), considering that the two functions are different?

 

[Qwertywerty]

I understand what both of you are saying, and it helps. But what I am asking is what guarantees that limx→1x+1=limx→1(x+1)(x−1)x−1 \displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1} (for example), considering that the two functions are different?

Different in the sense of domain only .

Although I mentioned this in my previous post -
Example - 5 × 10^-30 / 10^-30 = ?

 

[Mark44]

I understand what both of you are saying, and it helps. But what I am asking is what guarantees that $\displaystyle\lim_{x\rightarrow 1}x+1 = \displaystyle\lim_{x\rightarrow 1}\frac{(x+1)(x-1)}{x - 1}$ (for example), considering that the two functions are different?

The two functions are different at only a single point; namely, at (1, 2). At all other points they are identical.

$\lim_{x \to 1}\frac{(x+1)(x-1)}{x - 1} = \lim_{x \to 1}(x + 1) \cdot \lim_{x \to 1}\frac{x - 1}{x - 1}$, using the property of limits that the limit of a product is the product of the limits, provided that both limits exist.
The first limit in the product is clearly 2. What would you say that the second limit is?

 

[Mr Davis 97]

Different in the sense of domain only .

Although I mentioned this in my previous post -
Example - 5 × 10^-30 / 10^-30 = ?

I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do $\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}$, then $\displaystyle\lim_{x\rightarrow 0} 1$, the answer would be 1. But f(x) of the original problem is not the same as 1. So how do we know for sure that $\displaystyle\lim_{x\rightarrow 0} 1$ answers the original question of $\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}$?

 

[Mark44]

I still don't understand. Consider f(x) = x / x. This function is defined for all real numbers except 0. If we do $\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}$, then $\displaystyle\lim_{x\rightarrow 0} 1$, the answer would be 1. But f(x) of the original problem is not the same as 1.

Yes, it is, except at a single point. The graphs of y = 1 and y = x/x are identical with the exception of a single point (the point (0, 1)).

So how do we know for sure that $\displaystyle\lim_{x\rightarrow 0} 1$ answers the original question of $\displaystyle\lim_{x\rightarrow 0} \frac{x}{x}$?

 

[Mr Davis 97]

Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."

 

[Qwertywerty]

Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."

Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .

 

[micromass]

Just a side note - The value of say 0.99999 ... in your original question is 1.99999 ... , but at x = 1 , we would only say it should be approaching the value 2 .

And just a note: 0.99999... = 1, and 1.99999... = 2.

 

[FactChecker]

Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."

If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.

Keeping track of the valid domain is a good habit anyway.

 

[pbuk]

Okay, so tell me if this mental model is correct for evaluating limits algebraically. "I'm evaluating what value this function approaches as the argument approaches a certain value. If the function is smooth, then I know that it will approach the value that is defined at the point in question. If the function has a discontinuity at the point in question, then my job is to find some other function that is identical to the one in question except at that point, and evaluate at the point, thereby concluding that the original function must be approaching it."

No, I don't think that will work for example for $\lim_{x \to 0} \frac{\sin x}x$.

To find a limit, your job is to consider what happens as the argument approaches the limit. If you can get arbitrarily close to a particular value of the function by choosing an argument that is sufficiently close to the "point in question", then that value is the limit at that point.

Apart from the technique you mention I can think of two methods to find $\lim_{x \to a} f(x)$ that are often useful:

• find a function g(x) that is always further from the limit L than f(x) is from L and show that $\lim_{x \to a} g(x) = L$
• express f(x) as the sum of a number (often an infinite number) of terms most of which approach 0 as x->a

 

[jbriggs444]

But what justification do we have for cancelling the (x - 1) terms? When we cancel these terms, we are effectively dealing with a whole new function since the domain changes. Why is changing functions allowed in doing limits? How can we be absolutely certain that $\lim_{x\rightarrow 1}x+1$ leads to the correct answer to the original problem if $x + 2$ is different function, with a different domain, than $\frac{x^2 -1}{x - 1}$?

If you examine the standard epsilon/delta definition of limits, you will see that the value of the function at the limit point is completely irrelevant to the definition. Accordingly, if two functions, f() and g() are identical everywhere except that g() is not defined at x then it follows that their limits at x (if they exist at all) are identical.

 

[daqddyo1]

The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.

 

[jbriggs444]

The point about the meaning of limx→a is that x can approach as close as you like in value to a but it can never equal a. As a result, the difference between the value of the function at x=a whose limit is being determined and the value of the function at a value of x exceptionally close to a is infintesmal.

That only holds for continuous functions where, by definition of continuity, the value of the function at a point must be equal to the limit of the function at that point.

 

[Mr Davis 97]

If you are worried about the functions not being the same at x=1, then you should just restrict the domain to omit x=1. F(x) = (x-1)(x+1)/(x-1) and G(x) = x+1, x≠1 are identical functions.

Keeping track of the valid domain is a good habit anyway.

I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.

 

[micromass]

Theorem: If $f:D\rightarrow \mathbb{R}$ and $g:D\rightarrow \mathbb{R}$ are functions which agree everywhere except possibly in $a$, then $\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)$.

 

[FactChecker]

I would do this, except that I need to evaluate x + 1 at x = 1 in order to find the limit of the original problem. Therefore, it doesn't seem like I could exclude x = 1 from the second function's domain.

Not really. When you determine a limit, you avoid the final point in the domain. So the function can be undefined at that point. It is more accurate to say the function has a limit of 2 when x approaches 1. So I can say that f(x) = (x+1)(x-1)/(x-1) has a limit of 2 as x approaches 1. f(x) is undefined at x=1 unless I add f(1)=2 to the definition of f.

That is one difference between 'continuity' and 'limit'. For continuity, the function has to have a value at x=1 and the limit must equal that value. Suppose I define f(x)=(x+1)(x-1)/(x-1) for x≠1, f(1)=9999. Then the limit of f as x approaches 1 is 2, but the value of f(1) is 9999. So it has a limit but is not continuous,

 

[Mr Davis 97]

Theorem: If $f:D\rightarrow \mathbb{R}$ and $g:D\rightarrow \mathbb{R}$ are functions which agree everywhere except possibly in $a$, then $\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a} g(x)$.

Thank you! That's exactly what I was looking for.