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Finding the limit of a rational function

  1. Jun 22, 2012 #1
    1. The problem statement, all variables and given/known data

    limx→2(x+2)/(x^3+8)

    2.
    I only recently started learning calculus on my own so correct me if i'm wrong.
    When using direct substitution and the denominator equals 0, the limit is undefined, just like any fraction is when its denominator equals 0. However, it's limit can still be found through substitution by making f(x)=g(x) when x≠a.
    i.e limx→1 (x2-1)/(X-1) can be written as limx→1 (X+1) (canceling out by factorizing the numerator as a difference of squares). Through substitution we then find the limit to be 2.

    I also got the idea that when the numerator equals 0 through substitution the limit doesn't exist. i.e for limX→2 (x^2-x+6)/(x-2) no preliminary algebra exists to factor this and make it possible to find the limit through substitution so the limit doesn't exist.


    3. The attempt at a solution

    limx→2(x+2)/(x^3+8) doesn't appear to exist. However, on my text book it says it's -1/2...

    All help is appreciated! Let me know if anything i said is incorrect.
     
  2. jcsd
  3. Jun 22, 2012 #2
    Check that you have written down the problem correctly, because at present the denominator is not zero at the given "limit" and thus there is no need for anything but calculation.
     
  4. Jun 22, 2012 #3

    SammyS

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    First of all: There is no difficulty in taking the limit [itex]\displaystyle \lim_{x\to2}\frac{x+2}{x^3+8}\,.[/itex]

    That function is continuous at x=2, so it looks like you have a typo.

    By the Way: It is possible to factor the sum (and/or difference) of cubes.

    a3 + b3 = (a + b)(a2 - ab + b2)
     
  5. Jun 22, 2012 #4

    Mentallic

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    In the case where you had [tex]\lim_{x\to 1}\frac{x^2-1}{x-1}[/tex] Notice that x=1 is a root of both the numerator and denominator because if we plug x=1 into it, we get the indeterminate form 0/0, thus, (x-1) is a factor of both numerator and denominator. Factoring out and then cancelling will give us the desired limit that we can solve.

    But what exactly does it mean? Well, if you draw the graph of [tex]y=\frac{x}{x-1}[/tex], this is a hyperbole which has an asymptote at x=1. This means that [tex]\lim_{x\to 1}\frac{x}{x-1}[/tex] is undefined. In this case, when we plug x=1 into the equation, we get the form 1/0 as opposed to 0/0 as we did before.

    Well what does the graph [tex]y=\frac{x^2-1}{x-1}[/tex] look like? While we're dividing by a linear equation like we did in the case of [tex]y=\frac{x}{x-1}[/tex] since we can simplify it into x+1 by cancelling factors, it's actually the graph y=x+1 but it has a hole at x=1.

    Ok, so basically the moral of the story is if you have to solve a limit and when you plug some value x=a (a being some constant) into the equation you get the indeterminate form 0/0, then that means that you can factor out (x-a) from both the numerator and denominator and simplify it down. If you get the form 1/0 (or any non-zero constant divided by 0) then it's undefined.

    So what can you say about [tex]\lim_{x\to -2} \frac{x+2}{x^3+8}[/tex] :smile:

    Oh and be careful, [itex]x^2-x+6[/itex] can't be factored, yes, but you said that it is equal to zero when x=2. This is wrong. You have to keep in mind that if x=2 does make it equal to zero, then you can factor out x-2 from the quadratic, but since you can't factor it, x=2 does not give you a value of 0.
     
  6. Jun 22, 2012 #5
    Yes, there are two typos. it's the limx-2 for the question itself. and when i gave an example for a numerator = 0 i don't know why i gave a wrong one. but what ever, basically what i meant was that when the denominator equals 0 it's undefined and its the same idea for when the numerator equals 0...
     
    Last edited: Jun 22, 2012
  7. Jun 22, 2012 #6

    HallsofIvy

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    Yes, that's true. If f(x)= g(x) for all x except a, we still have [itex]\lim_{x\to a} f(x)= \lim_{x\to a} g(x)[/itex].

    No, when the denominator equals 0, direct substitution cannot be used but the limit itself may or may not exist.

    In this example, at x= 2 the denominator is 0 while the numerator is not. As x approches 0, the denominator approaches 0 while the numerator approaches 8. The fraction gets larger and larger without bound.

    For the problem you give the denominator goes to 16, not 0. The numerator goes to 4 so the limit is 4/16= 1/4.

    If the problem was, in fact, the limit as x goes to -2, then you need to use the fact that [itex]x^3+ 8= (x+ 2)(x^2+ 2x+ 4)[/itex]. Now, the (x+ 2) terms cancel and you can take the limit as of [itex]1/(x^2+ 2x+ 4)[/itex] which is [itex]1/(4+ 4+ 4)= 1/12[/itex]. So either your textbook has the wrong problem or you have copied it incorrectly.

    By the way, in general, if a makes a poynomial equal to 0, if P(a)= 0, then x- a is a factor.
     
    Last edited: Jun 23, 2012
  8. Jun 22, 2012 #7

    HallsofIvy

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    "x-2" is meaningless. Is it "[itex]x\to 2[/itex]" or [itex]x\to -2[/itex]? In either case, as I said above, the limit exists but is NOT equal to -1/2.
     
  9. Jun 22, 2012 #8
    No way, Jose !
    Hi, I think you got it wrong, as has been suggested before, you may have a few typos.
    Didn't you mean limit as x goes to -2 instead of 2 ?
    I suppose this is what you meant, and you can use De l'Hopital's rule if you know about it (if not, well, then not), or (much better anyway) you can see that
    x³+8=x³+2³=(x+2)(x²-2x+2²) therefore you can simplify by x+2 and your limit, as you would find with de l'hopital's rule would be 1/12 (not -1/2, another likely typo)

    Cheers...
     
  10. Jun 22, 2012 #9
    The only interesting case is when the numerator and denominator become zero at the same value. If the numerator is non-zero when the denominator is zero, the function does not have a finite value. If the denominator is non-zero when the numerator is zero, of course the function value is zero. When they both go to zero together - like say ##\frac{3x^5}{x^5}## - then you can think about limit values. Whether you do this by factoring or by other means is another question.
     
  11. Jun 22, 2012 #10

    SammyS

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    Let's see if OP returns.
     
  12. Jun 23, 2012 #11
    I've understood everything. Thank you all for your feedback... basically when we get the indeterminate form 0/0, preliminary algebra can be used to find another fuction equal to f(x) when x≠a. When the numerator is zero, then there's obviously no limit. The only doubt i have is when the denominator is zero. If the denominator is zero, there's no finite limit. Does this mean I will always get a asymptote were x=a?
     
  13. Jun 23, 2012 #12

    Mentallic

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    No! When the numerator is zero and your denominator is non-zero, then the limit is zero!

    Yes, exactly. But the answer isn't infinite either, it's undefined.

    Think about [tex]\lim_{x\to 0}\frac{1}{x}[/tex] the limit tends to infinite when 0 is approached from the positive side, but it approaches negative infinite when approached from the negative side. We denote this as

    [tex]\lim_{x\to 0^+}\frac{1}{x}=+\infty[/tex]
    [tex]\lim_{x\to 0^-}\frac{1}{x}=-\infty[/tex]

    For the limit to exist, both sides of the limit must tend towards the same value, and hence since this isn't the case, it's undefined.
     
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