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Math Help - Limit...using given info

  1. #1
    Senior Member polymerase's Avatar
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    Limit...using given info

    Given that the tangent line to the graph of f(x) at (0,0) has equation 2y=x and that f(x) has a horizontal asymptote at \infty with equation y=2, find the value of

    \displaystyle\lim_{x\to{0^{+}}} {\frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x})
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Given that the tangent line to the graph of f(x) at (0,0) has equation 2y=x and that f(x) has a horizontal asymptote at \infty with equation y=2, find the value of

    \displaystyle\lim_{x\to{0^{+}}} {\frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x})
    I'm not any sort of an expert at limits, but this looks so I'm going to take a shot at it.
    First since \lim_{x \to +\infty}f(x) = 2 we know that \lim_{x \to 0^+}f \left ( \frac{1}{x} \right ) = 2

    Thus
    \lim_{x \to 0+}x^2 f \left ( \frac{1}{x} \right ) = \lim_{x \to 0^+} 2x^2 = 0

    Next we have \lim_{x \to 0^+} \frac{sin(2x)}{f(x)}. We know that \lim_{x \to 0}f(x) = 0 since the tangent line to y = f(x) at x = 0 is 2y = x. Thus the limit is of the form \frac{0}{0} and we may apply L'Hopital's rule:
    \lim_{x \to 0^+} \frac{sin(2x)}{f(x)} = \lim_{x \to 0^+} \frac{2 cos(2x)}{\frac{1}{2}} = \lim_{x \to 0^+}4cos(2x) = 4
    (Because f(x) \approx \frac{1}{2}x near x = 0.)

    Thus
    \lim_{x \to 0^+} { \frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x}) = 4 - 0 = 4

    -Dan
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    I'm not any sort of an expert at limits, but this looks so I'm going to take a shot at it.
    First since \lim_{x \to +\infty}f(x) = 2 we know that \lim_{x \to 0^+}f \left ( \frac{1}{x} \right ) = 2

    Thus
    \lim_{x \to 0+}x^2 f \left ( \frac{1}{x} \right ) = \lim_{x \to 0^+} 2x^2 = 0

    Next we have \lim_{x \to 0^+} \frac{sin(2x)}{f(x)}. We know that \lim_{x \to 0}f(x) = 0 since the tangent line to y = f(x) at x = 0 is 2y = x. Thus the limit is of the form \frac{0}{0} and we may apply L'Hopital's rule:
    \lim_{x \to 0^+} \frac{sin(2x)}{f(x)} = \lim_{x \to 0^+} \frac{2 cos(2x)}{\frac{1}{2}} = \lim_{x \to 0^+}4cos(2x) = 4
    (Because f(x) \approx \frac{1}{2}x near x = 0.)

    Thus
    \lim_{x \to 0^+} { \frac{sin(2x)}{f(x)}} - x^2f(\frac{1}{x}) = 4 - 0 = 4

    -Dan
    Thanks!
    By the way, you're right .
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    Thanks!
    By the way, you're right .


    That was fun! Thank you for the problem.

    -Dan
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post


    That was fun! Thank you for the problem.

    -Dan
    You think you had a hard time...i had to do this problem for my calculus term test.
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