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Math Help - Calculus 6e James Stewart

  1. #1
    Member SengNee's Avatar
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    Calculus 6e James Stewart

    Calculus 6th Edition
    Metric International Edition
    James Stewart

    page 86.

    55)
    If \lim_{x\rightarrow 1}\frac{f(x)-8}{x-1}=10, find \lim_{x\rightarrow 1}f(x).


    57)
    If f(x)=x^2 if x is rational.
    =0 if x is irrational.

    prove that \lim_{x\rightarrow 0}f(x)=0.


    59)
    Show by means of an example that \lim_{x\rightarrow a}[f(x)g(x)] may exist even though neither \lim_{x\rightarrow a}f(x) nor \lim_{x\rightarrow a}g(x) exists.


    61)
    Is there a number a such that

    \lim_{x\rightarrow -2}\frac{3x^2+ax+a+3}{x^2+x-2}

    exist? If so, find the value of a and the value of the limit.
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  2. #2
    MHF Contributor matheagle's Avatar
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    The first one should be 8 since you want 0/0 here.

    I would use the squeeze/turkey sandwich theorem here.
    f is either the square or 0. Hence

    0\le |f(x)-0|\le x^2\to 0 as  x\to 0.

    Third one... g(x)=f(x)=(-1)^x so g(x)f(x)=1 for all x.
    Here's I'm thinking of x as a natural number heading towards infinity, but I don't think it matters where we head.

    Iin the last one we need
    3(-2)^2+a(-2)+a+3=0, which seems to be 15.
    It's 2am and I'm done for tonight.

    With a=2 we have

     \lim_{x\to -2}{3(x^2+5x+6)\over x^2+x-2}=\lim_{x\to -2}{3(x+3)(x+2)\over (x+2)(x-1)}

     =3\lim_{x\to -2}{x+3\over x-1}..............
    Last edited by matheagle; May 26th 2009 at 11:57 PM.
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  3. #3
    Member SengNee's Avatar
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    Quote Originally Posted by matheagle View Post
    The first one should be 8 since you wnat 0/0 here.
    Why I should make it becomes 0/0?

    Quote Originally Posted by matheagle View Post
    Third one... g(x)=f(x)=(-1)^x so g(x)f(x)=1 for all x.
    Here's I'm thinking of x as a natural number heading towards infinity, but I don't think it matters where we head.
    Is it means for all x in the domain(Natural numbers)? If yes, the function will be discontinuous, then how can it be differentiated?

    Quote Originally Posted by matheagle View Post
    the last one we need
    3(-2)^2+a(-2)+a+3=0, which seems to be 15.
    It's 2am and I'm done for tonight.

    With a=2 we have

     \lim_{x\to -2}{3(x^2+5x+6)\over x^2+x-2}=\lim_{x\to -2}{3(x+3)(x+2)\over (x+2)(x-1)}

     =3\lim_{x\to -2}{x+3\over x-1}..............
    Again, why should I make it 0/0?
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  4. #4
    MHF Contributor matheagle's Avatar
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    The denomiator is heading towards zero.
    If you want the limit to be 8 the numerator must also be heading towards zero.
    Same comment on the last one, if the numerator is heading towards, say 5, and the denominator
    is heading towards zero, then you either have a plus or minus infinity, or a does not exist situation.
    Last edited by matheagle; May 27th 2009 at 08:22 AM.
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