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Thread: Limits question

  1. #1
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    Limits question

    Find\ all\ (a,b)\ such\ that

    \lim_{x \rightarrow 0} \frac{ax-1+e^{bx}}{x^2}=1

    I can find b by using L'Hopital's Rule, where b is \pm \sqrt{2}, but I don't know how to find a
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Find\ all\ (a,b)\ such\ that

    \lim_{x \rightarrow 0} \frac{ax-1+e^{bx}}{x^2}=1

    I can find b by using L'Hopital's Rule, where b is \pm \sqrt{2}, but I don't know how to find a
    There is no restriction on what a can be. a is an element of R. This is clear when you use L'Hopital twice.
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  3. #3
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    Quote Originally Posted by Debsta View Post
    There is no restriction on what a can be. a is an element of R. This is clear when you use L'Hopital twice.
    But the answer says,

    (a,b)=(-\sqrt{2}, \sqrt{2})\ or\ (\sqrt{2}, -\sqrt{2})
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Find\ all\ (a,b)\ such\ that

    \lim_{x \rightarrow 0} \frac{ax-1+e^{bx}}{x^2}=1

    I can find b by using L'Hopital's Rule, where b is \pm \sqrt{2}, but I don't know how to find a
    I'm not clear how you can use L'Hopital and NOT get a. Differentiating both numerator and denominator, gives \frac{a+ be^{bx}}{2x} and using L'Hopital's rule again, \frac{b^2e^{bx}}{2} which has a limit of \frac{b^2}{2}. In order for that to be 1, b must be \pm\sqrt{2}.

    But, in order that we be able to use L'Hopital a second time, on the fraction \frac{a+ be^{bx}}{2x}, since the denominator goes to 0, the numerator must also- we must have \lim_{x\to 0}a+ be^{bx}= a+ b= 0. That is a= -b.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    I'm not clear how you can use L'Hopital and NOT get a. Differentiating both numerator and denominator, gives \frac{a+ be^{bx}}{2x} and using L'Hopital's rule again, \frac{b^2e^{bx}}{2} which has a limit of \frac{b^2}{2}. In order for that to be 1, b must be \pm\sqrt{2}.

    But, in order that we be able to use L'Hopital a second time, on the fraction \frac{a+ be^{bx}}{2x}, since the denominator goes to 0, the numerator must also- we must have \lim_{x\to 0}a+ be^{bx}= a+ b= 0. That is a= -b.
    Yes my mistake. I overlooked the fact that we must have 0/0 for L'Hopital's rule to apply. Sorry for the confusion.
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