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Functions and their domains

  1. Feb 12, 2009 #1
    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.
  2. jcsd
  3. Feb 12, 2009 #2


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    The point you are making is correct, but it is quibbling.
  4. Feb 12, 2009 #3


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    [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.
  5. Feb 12, 2009 #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?
  6. Feb 12, 2009 #5


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    I don't consider it quibbling! Any good text book 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]
    [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.
  7. Feb 12, 2009 #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.
  8. Feb 12, 2009 #7

    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?
  9. Feb 12, 2009 #8
    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.

    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.
  10. Feb 12, 2009 #9
    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.
  11. Feb 12, 2009 #10
    Ah. Gotcha.
  12. Feb 12, 2009 #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).
  13. Feb 12, 2009 #12


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    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
    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
    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
    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]
  14. Feb 13, 2009 #13


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    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.
  15. Feb 13, 2009 #14
    Thanks Halls, that cleared everything up :)
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