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Math Help - Calculate the limit (Question 3)

  1. #1
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    Calculate the limit (Question 3)

    I am stuck with the following question and can't finish it:

    \text{Calculate }\lim_{x \rightarrow 4 } \frac{4x-x^2}{2-\sqrt x}

    = \frac{4(4)-4^2}{2-\sqrt4}

    = \frac{0}{0 }\text{Indeterminant}

    = \lim_{x \rightarrow 4 } \frac{4x-x^2}{2-\sqrt x} \cdot \frac{2+\sqrt x}{2+\sqrt x}

    = (2-\sqrt x) \cdot (2+ \sqrt x)

    = 4-x

    = \lim_{x \rightarrow 4 } \frac{(4x-x^2) \cdot (2+ \sqrt x)}{4-x} Something looks wrong here! Where did I go wrong?
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  2. #2
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    Re: Calculate the limit (Question 3)

    Nothing is wrong; you're so close!

    You simply need to factor the (4x-x^2) into x(4-x)

    (4-x) cancels on top and bottom, and then you just plug in the 4.

    -Nathan
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  3. #3
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    Re: Calculate the limit (Question 3)

    Quote Originally Posted by TutorMeNate View Post
    Nothing is wrong; you're so close!

    You simply need to factor the (4x-x^2) into x(4-x)

    (4-x) cancels on top and bottom, and then you just plug in the 4.

    -Nathan
    Yes! I did not see that. Ok

    \lim_{x \rightarrow 4 } \frac{(4x-x^2) \cdot (2+ \sqrt x)}{4-x}

    = \lim_{x \rightarrow 4 } \frac{(4-x)(x) \cdot (2+ \sqrt x)}{4-x}

    = x(2+ \sqrt x)

    = 4(2+ \sqrt 4)

    = 16

    Is this correct?
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  4. #4
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    Re: Calculate the limit (Question 3)

    Yes, perfect!

    -Nathan
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  5. #5
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    Re: Calculate the limit (Question 3)

    Quote Originally Posted by TutorMeNate View Post
    Yes, perfect!

    -Nathan
    Nice, thanks again!
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