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Math Help - [SOLVED] Limit question

  1. #1
    PTL
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    [SOLVED] Limit question

    limit as h goes to 0 of  \frac{f(2+h)-f(2)}{h} where f(x) = \sqrt{x}

    It looks similar to differentiation from first principles, but I can't see how to use that...
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  2. #2
    Moo
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    Hello,

    Remember that (a-b)(a+b)=a^2-b^2

    This is a common trick : multiply by 1=\frac{f(2+h)+f(2)}{f(2+h)+f(2)}

    The numerator will simplify with the denominator, and what's left is no longer undefined
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  3. #3
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    Quote Originally Posted by PTL View Post
    limit as h goes to 0 of  \frac{f(2+h)-f(2)}{h} where f(x) = \sqrt{x}

    It looks similar to differentiation from first principles, but I can't see how to use that...
    \lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h}

    Rationalize the numerator:
    = \lim_{h \to 0} \frac{\sqrt{2 + h} - \sqrt{2}}{h} \cdot \frac{\sqrt{2 + h} + \sqrt{2}}{\sqrt{2 + h} + \sqrt{2}}

    = \lim_{h \to 0} \frac{2 + h - 2}{h(\sqrt{2 + h} + \sqrt{2})}

    = \lim_{h \to 0} \frac{h}{h(\sqrt{2 + h} + \sqrt{2})}

    = \lim_{h \to 0} \frac{1}{\sqrt{2 + h} + \sqrt{2}}

    Now you can evaluate the limit by direct substitution.


    01


    EDIT: too slow...
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by PTL View Post
    limit as h goes to 0 of  \frac{f(2+h)-f(2)}{h} where f(x) = \sqrt{x}

    It looks similar to differentiation from first principles, but I can't see how to use that...
    If you are supposed to know that f is differentiable at x=2 then you can claim that
     \lim_{h\to0}\frac{f(2+h)-f(2)}{h}=f'(2).
    and all you have to do is compute f'(2)...
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