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Math Help - Limits problems without #'s

  1. #1
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    Limits problems without #'s

    Does this problem make anyones else's eyes bug out?

    If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
    h->0


    b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
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  2. #2
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    But Dan, we all know that if f(x)=\sqrt x\implies f'(x)=\frac1{2\sqrt x}

    But he's asking for f(x)=\frac1{\sqrt x} which is f'(x)=2\sqrt x
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  3. #3
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    Wow! You make it seem so easy! Thanks a bunch!

    [EDIT] So wait...then the answer is 2√x?
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  4. #4
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    Quote Originally Posted by KennyYang View Post
    Does this problem make anyones else's eyes bug out?

    If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
    h->0


    b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
    Krizalid isn't quite right here.

    Round 2!

    f(x) = \frac{1}{\sqrt{x}}

    f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}

    Simplify a bit first:
    f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \cdot \frac{\sqrt{x + h} \sqrt{x} }{\sqrt{x + h} \sqrt{x} }

    = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}}

    Again, rationalize the numerator:
    = \lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h} \sqrt{x}} \cdot \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}}

    = \lim_{h \to 0} \frac{x - (x + h)}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}

    = \lim_{h \to 0} \frac{-h}{h\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}

    = -\lim_{h \to 0} \frac{1}{\sqrt{x + h} \sqrt{x}(\sqrt{x} + \sqrt{x + h})}

    = -\frac{1}{\sqrt{x} \sqrt{x}(\sqrt{x} + \sqrt{x})}

    = -\frac{1}{x(2\sqrt{x})}

    = -\frac{1}{2x \sqrt{x}}

    Which is typically written as -\frac{1}{2}x^{-3/2}

    -Dan
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    But he's asking for f(x)=\frac1{\sqrt x} which is f'(x)=2\sqrt x
    Oops!, I made a mistake here , I confused the derivative for an integral
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Oops!, I made a mistake here , I confused the derivative for an integral
    It's okay. We still love you. (In a platonic way, of course! )

    -Dan
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  7. #7
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    I gave part (b) a shot and came up with such a convoluted answer that I'm not going to bother to post it. Did anyone else get a ridiculously overcomplicated answer as well?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by KennyYang View Post
    b. If the above limit represents a slope function for the line tangent to f at (x, f(x)). Write the equation of the tangent line at x=4, in point-slope form.
    Quote Originally Posted by topsquark View Post
    f^{\prime}(x) = -\frac{1}{2}x^{-3/2}

    -Dan
    Any line can be expressed in the form y = mx + b. The slope of the tangent line at a point (x, f(x)) of the graph is f^{\prime}(x), so the slope of the tangent line is, in this case:
    m = f^{\prime}(4) = -\frac{1}{2 \cdot 4^{3/2}} = - \frac{1}{2 \cdot 8} = -\frac{1}{16}

    So we have the form:
    y = -\frac{1}{16}x + b

    Now, we know that the function f(x) and the tangent line have the same values at the point x = 4. So we know that
    f(4) = y(4) = -\frac{1}{16} \cdot 4 + b

    f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} = <br />
-\frac{1}{16} \cdot 4 + b

    b = \frac{1}{2} + \frac{1}{16} \cdot 4 = \frac{3}{4}

    So I get that the tangent line to f(x) at x = 4 is y = -\frac{1}{16}x + \frac{3}{4}.

    -Dan
    Attached Thumbnails Attached Thumbnails Limits problems without #'s-tangent.jpg  
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    Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by KennyYang View Post
    Guess it was just a mathematical error on my part. Thanks for all your help today topsquark!
    No problem!

    -Dan
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  11. #11
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    Quote Originally Posted by KennyYang View Post
    If f(x) = 1/(√x) find lim [f(x+h)-f(x)]/ h
    h->0
    In fact the limit is the derivative of f(x) at x.
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