Understanding Function Equality and Domain Restrictions

In summary, the conversation discusses the concept of two functions being equal to each other if their equations are the same and their domains are the same. However, this is not always the case as seen in the example of f(x) = \frac{x^2-1}{x-1} and g(x) = x+1, where f(x) is not defined at x = 1 but g(x) is. The limit of these two functions at x = 1 is the same due to the rules of calculus, but they are not the same function. The concept of limits is independent of the graph of the function and is used to find the "real" limit.
  • #1
JG89
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Suppose [tex] f(x) = \frac{x^2 - 1}{x-1} [/tex]. Why do we say that [tex] f(x) = \frac{x^2 - 1}{x-1} = x + 1 [/tex], if [tex]\frac{x^2 - 1}{x-1}[/tex] isn't defined at x = 1, but (x + 1) is defined at x = 1.

I always thought that we say two functions are equal to each other if their equations are the same and their domain is the same.
 
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  • #2
The point you are making is correct, but it is quibbling.
 
  • #3
[tex] f(x) = \frac{x^2 - 1}{x-1} [/tex] and [tex] f(x) = x + 1 [/tex] are not the same function. The first one contains a factor of 1, made of a binomial. The first function must exclude x=1, but the second function includes x=1. When you simplify the first kind of function, you loose a factor (in this case which contains the independant variable), which siginficantly can change the domain.
 
  • #4
Using those same functions, suppose we wanted to find the limit as x tends to 1. Why are we allowed to pass to the limit in (x+1) but not in the first function? I know it may be because the second function is continuous, but how do we know that passing to the limit in (x+1) is the "real" limit of the first function?
 
  • #5
I don't consider it quibbling! Any good textbook will state
[itex]\frac{x^2- 1}{x-1}= x+ 1[/tex] for x not equal to 1

Note, however, that it is true that
[tex]\lim_{x\rightarrow 1}\frac{x^2- 1}{x- 1}= \lim_{x\rightarrow 1} x+ 1= 2[/tex]
because
[tex]\lim_{x\rightarrow x_0} f(x)[/tex]
is independent of the value of [itex]f(x_0)[/itex] and may exist even when [itex]f(x_0)[/itex] doesn't.
 
  • #6
Well if you claim that [tex] f(x) = \frac{x^2-1}{x-1} [/tex] is a function that implies that 1 is NOT in the domain since by definition of a function, all elements of the domain map somewhere.
 
  • #7
HallsofIvy said:
I don't consider it quibbling! Any good textbook will state
[itex]\frac{x^2- 1}{x-1}= x+ 1[/tex] for x not equal to 1

Note, however, that it is true that
[tex]\lim_{x\rightarrow 1}\frac{x^2- 1}{x- 1}= \lim_{x\rightarrow 1} x+ 1= 2[/tex]
because
[tex]\lim_{x\rightarrow x_0} f(x)[/tex]
is independent of the value of [itex]f(x_0)[/itex] and may exist even when [itex]f(x_0)[/itex] doesn't.


How can we guarantee that the limits are the same all the time? I mean, in this specific example if you graph both functions, you can see that at (x^2 - 1)/(x-1) there is a hole at x = 1 and that the x values approach 2, and it is easy to see that the limit of (x+1) is the same, but how do we know this is so in all possible cases?
 
  • #8
NoMoreExams said:
Well if you claim that [tex] f(x) = \frac{x^2-1}{x-1} [/tex] is a function that implies that 1 is NOT in the domain since by definition of a function, all elements of the domain map somewhere.

The domain of f is R/{1}. It is clear that f *does* map all points in its domain to points in its codomain. It is a function and there is no problem with calling it one, as long as you are careful about the restrictions on its domain.

JG89 said:
How can we guarantee that the limits are the same all the time? I mean, in this specific example if you graph both functions, you can see that at (x^2 - 1)/(x-1) there is a hole at x = 1 and that the x values approach 2, and it is easy to see that the limit of (x+1) is the same, but how do we know this is so in all possible cases?

Limits have nothing to do with the graph of the function. The graph is simply a helpful tool to create an intuition of the graph's behavior.

I'm not sure what part you find confusing here, but I'll walk through it.

For all x, [tex]\frac{x^2 - 1}{x-1} = \frac{(x+1)(x-1)}{x-1}[/tex]. As long as [tex]x /= 1[/tex], we see the denominator is some nonzero number, so we can cancel out the [tex]x-1[/tex] from the top and the bottom, leaving [tex]x+1[/tex]. What we've just shown is that for all x other than 1, [tex]\frac{x^2 - 1}{x-1} = x+1[/tex].

Now, these two functions, [tex]\frac{x^2 - 1}{x-1}[/tex] and [tex]x+1[/tex] are NOT equal, because one is defined at 1 and the other isn't. However, we can show their limits are the same at 1 using rules from calculus. Under the "lim" operator, you are allowed to divide by polynomials even when they could potentially take on 0 as a value. That is, you can cancel the x-1 from the top with the x-1 from the bottom and say [tex]\lim \frac{x^2 - 1}{x-1} = \lim \frac{(x+1)(x-1)}{x-1} = x+1[/tex].

The reasoning behind why this is a legitimate operation is something that doesn't get fully explained until college-level analysis. However, you can imagine the limit operation as taking any number that is "infinitely" close to the number being approached. So if our limit is 1, then we take [tex]1 + \epsilon[/tex] for some extremely small number [tex]\epsilon[/tex]. So small, in fact, that it doesn't have any affect on the rest of the equation, and after dividing [tex]\frac{\epsilon}{\epsilon} = 1[/tex], we simply "round" [tex]\epsilon[/tex] to 0 in the remaining calculation.
 
  • #9
Tac-Tics said:
The domain of f is R/{1}. It is clear that f *does* map all points in its domain to points in its codomain. It is a function and there is no problem with calling it one, as long as you are careful about the restrictions on its domain.

I was responding to the OP saying that he thought they were the same function if the domains were the same, etc. and pointing out that the domains are not the same.
 
  • #10
NoMoreExams said:
I was responding to the OP saying that he thought they were the same function if the domains were the same, etc. and pointing out that the domains are not the same.

Ah. Gotcha.
 
  • #11
Tac-Tics, what you said was a little confusing. Here is my reasoning (for our example we've been using throughout the thread), and it would be great if you could tell me if it's correct or not:

Let h(x) = f(x)g(x) where [tex] f(x) = x + 1 [/tex] and [tex] g(x) = \frac{x-1}{x-1} [/tex]. f(x) is continuous and has a limit of 2 as x approaches 1, so for x sufficiently close to 1, we have [tex] |f(x) - 2| < \epsilon [/tex] for all epsilon greater than 0. Now, notice that g(x) is really equal to 1, which is continuous everywhere on the x-axis and has a limit of 1 for x approaching any point. So we have [tex] |g(x) - 1| < \epsilon* [/tex] for all x lying sufficiently close to 1, and where [tex] \epsilon \le \epsilon* [/tex]. Since the limits of f(x) and g(x) both exist at x = 1, then we can use the rule: "the limit of a the product is the product of its limit" and so we have:

[tex] |f(x)g(x) - 2| < \epsilon*[/tex], and thus the limit of h(x) = f(x)g(x) = 2 as x tends towards 1.

So in general, it is easy to see, using this example as a model, how a rational function where two factors cancel out, have the same limit as the factor that remains after the cancellation (assuming the limit exists in the first place).
 
  • #12
JG89 said:
Suppose [tex] f(x) = \frac{x^2 - 1}{x-1} [/tex]. Why do we say that [tex] f(x) = \frac{x^2 - 1}{x-1} = x + 1 [/tex], if [tex]\frac{x^2 - 1}{x-1}[/tex] isn't defined at x = 1, but (x + 1) is defined at x = 1.

I always thought that we say two functions are equal to each other if their equations are the same and their domain is the same.
The equations do not need to be the same
x+x and 2x are two equations that define the same function


To define a function one gives a domain and a rule
f(x)=x+1
is not a funtion because no domain has be given
often one fixes some set like the real numbers and considers functions with domain assumed to be all sensible real numbers
(x^2-1)/(x-1)=x+1
is true for all real numbers except 1 so those functions are equal so long as 1 is excluded from the domain.


when one defines a function by combining other functions some problems can arise
at each value both functions and their combinatn must exist
often as you point out potentially desired values can be excluded from a domain and there are several common ways to recover such values
1)Define the function piecewise
f(x)=sin(x)/x (x!=0)
f(x)=1 x=0
2)use weakened equivelence
f(x)=sin(x)/x
because I say so
3)use limits
f(x)=lim sin(x)/x
4)avoid problem
[tex]f(x)=\int_0^1 \cos(x t) dt[/tex]
 
  • #13
JG89 said:
How can we guarantee that the limits are the same all the time? I mean, in this specific example if you graph both functions, you can see that at (x^2 - 1)/(x-1) there is a hole at x = 1 and that the x values approach 2, and it is easy to see that the limit of (x+1) is the same, but how do we know this is so in all possible cases?

The definition of limit:
[tex]\lim_{x\rightarrow a} f(x)= L[/itex] if and only if, given any [tex]\epsilon> 0[tex] there exist [tex]\delta> 0[/tex] such that if [tex]0< |x- a|< \delta[/tex] then [tex]|f(x)- L|< \delta[/tex]

Notice that "0< |x- a|" part. As long as two functions are the same everwhere except at x= a, their limits, as x goes to a, must be the same, we never look at "x- a= 0" which is x= a. You really could calculate derivatives without that.

Of course, it is also true that it doesn't matter what the value is for [tex]|x-a|> \delta[/itex] for ANY positive [tex]\delta[/tex]. As long as f(x)= g(x) in some tiny "punctured" neighbor of a ("punctured" because a itself is removed) their limits as x goes to a are the same.

What is the limit, as x goes to 0 of

f(x)= 1000x100 if x< -0.00000000001
f(x)= x if -0.00000000001[itex]\le[itex] x< 0
f(0)= -10000000
f(x)= x if 0< x[itex]le[/itex] 0.00000000001
f(x)= e10000x if x> 0.00000000001 ?


0, of course, In the "punctured" neighborhood (- 0.00000000001, 0)U(0, 0.00000000001) f(x)= x.
 
  • #14
Thanks Halls, that cleared everything up :)
 

1. What is a function?

A function is a mathematical relationship that maps each element of a set (called the domain) to a unique element in another set (called the codomain).

2. What is the domain of a function?

The domain of a function is the set of all input values for which the function is defined. In other words, it is the set of all possible x-values that can be plugged into the function to produce an output.

3. How do you determine the domain of a function?

To determine the domain of a function, you need to look for any restrictions or limitations on the input values. Common restrictions include: division by zero, taking the square root of a negative number, and the presence of an even root in a radical expression.

4. Can a function have more than one domain?

No, a function can only have one domain. This is because each input value in the domain must correspond to exactly one output value in the codomain.

5. What is the difference between the domain and the range of a function?

The domain of a function is the set of all possible input values, while the range is the set of all possible output values. In other words, the domain is the set of x-values and the range is the set of y-values.

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