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Thread: limits w/ 0 as denominator

  1. #1

    Unhappy limits w/ 0 as denominator

    lim (x^2-3x+2)/(x^2-2x+1)

    i've tried factoring and cancelling but it still gives me a denominator of 0. i worked it out to...

    lim (x-2)/(x-1)

    ...and got a dead end. then i tried to do some research and found a question much like the above in the text:

    lim (x^2-x+6)/(x-2)

    the solution says though that the limit does not exist since x-2->0 (denominator) but x^2-x+6->8 (nominator) as x->2.

    that solution is saying that since the nominator and denominator don't agree, the limit doesn't exist.

    now back to my first problem using this theory. both the nominator and denominator approach 0 as x->1 so is it correct to say that the limit is therefore 0?

    lim (x^2-3x+2)/(x^2-2x+1) = 0??????????

    or did my deadend mean the limit does not exist??

    i'd really appreciate any help on this! ... that is if i worded everything so that it's comprehensible.
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  2. #2
    Senior Member
    Jun 2005
    For values of x not equal to 1, we have (x^2-3x+2)/(x^2-2x+1) = (x-2)(x-1)/(x-1)^2 = (x-2)/(x-1). This has no limit as x tends to 1: more precisely, as x tends to 1 from below, written x -> 1-, it tends to +infinity (that is, increases indefinitely) and as x tends to 1 from above, x -> 1+, it tends to -infinity (increases indefinitely in magnitude but with negative sign).

    Given a function N/D, if each of N and D tend to a limit as x->a, say n and d respectively, then you can say: if d is non-zero, N/D -> n/d; if d=0 but n non-zero then N/D does not tend to a limit; if d=0 and n=0 then you can't tell immediately.

    There's a useful further rule in the case when N and D both tend to zero: L'Hopital's rule. This says that if N and D are differentiable, N and D both -> 0 and the derivatives N' and D' tend to limits, then N/D has the same limiting behaviour as N'/D'. In your case N' is 2x-3 and D' is 2x-1. As x->1, we have N'->-1, D'->0 showing that there is no limit for N'/D' and so no limit for N/D.
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  3. #3
    Nov 2005
    that solution is saying that since the nominator and denominator don't agree, the limit doesn't exist.
    The correct word is numerator; anyway, what it is saying is that if the denominator goes to 0 and the numerator does not go to 0, the limit does not exist (more precisely, it "goes to infinity"). However, if the numerator went to 6 and the denominator went to 3, then the limit would be 2.
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  4. #4
    Global Moderator

    Nov 2005
    New York City
    Use L'Hopitals rule if it gives the form 0/0. Or Bring it to the form 0/0, but I do not know if your professor taught it to you yet.
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  5. #5
    Nov 2005
    There are some tricky things that can be done...

    For example, if you get something of the form 0^0, you can change it to something of the form e^(0*ln(0))=e^(0*-infinity) and use L'Hopital's Rule there. Also, if you have something of the form infinity/infinity, you can use L'Hopital's rule directly.

    I shall now attempt to use this trick to prove that if f and g are analytic real-valued functions of a real variable not everywhere 0 that vanish at the origin, lim(f(x)^g(x),x->0)=1.
    First, express this as e^(g(x)*ln(f(x))); then use L'Hopital's rule to obtain
    To simplify things, I shall assume that their derivatives are nonzero at the origin; a complete proof is found in Louis M. Rotando and Henry Korn, The Indeterminate Form 0^0, Mathematics Magazine, January 1977, p. 41.
    Then the aforementioned limit involves e^(-f'(0)/g'(0)), which I shall call A. The limit is the same as that of A^(g(x)^2/f(x)), and using L'Hopital's Rule again yields A^(2*g(x)*g'(x)/f'(x)), which goes to A^(2*0*g'(0)/f'(0))=1.
    A refinement would involve using more derivatives and uses of L'Hopital's Rule, but only finitely many; for if all of the derivatives of an analytic function are 0, the function is the zero function.
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