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Thread: Evaluate the Limits

  1. #1
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    Evaluate the Limits

    Question: If $\displaystyle f(9) = 9 , f'(9) = 4$ ,evaluate $\displaystyle \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$
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    Quote Originally Posted by zorro View Post
    Question: If $\displaystyle f(9) = 9 , f'(9) = 4$ ,evaluate $\displaystyle \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$

    As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.

    Tonio
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    Is this correct?

    Quote Originally Posted by tonio View Post
    As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.

    Tonio
    Here is what i have done



    $\displaystyle \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$


    Using L'Hospitals rule

    $\displaystyle g'(x)$ = $\displaystyle \frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\displaystyle \frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)

    $\displaystyle h'(x)$ = $\displaystyle \frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\displaystyle \frac{1}{a \sqrt{x}}$ ..................eq(2)

    Now taking lim of eq (1) and eq(2) we get

    $\displaystyle \lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\displaystyle \lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\displaystyle \frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $\displaystyle 4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
    Last edited by zorro; Dec 27th 2009 at 07:09 PM.
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    Quote Originally Posted by zorro View Post
    Here is what i have done

    $\displaystyle \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$


    Using L'Hospitals rule

    $\displaystyle g'(x)$ = $\displaystyle \frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\displaystyle \frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)

    $\displaystyle h'(x)$ = $\displaystyle \frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\displaystyle \frac{1}{a \sqrt{x}}$ ..................eq(2)

    Now taking lim of eq (1) and eq(2) we get

    $\displaystyle \lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\displaystyle \lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\displaystyle \frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $\displaystyle 4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
    Yes.
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