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Math Help - Find the Limit of x--->6

  1. #1
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    Find the Limit of x--->6

    Can someone kindly help me with this math problem for my calculus class.

    I'm sorry I do not have anything to post the proper form of the question (such as square root symbols) but i will try to do my best.

    Find lim x--> 6 [Squareroot(x-2) - 2] / (x-6) note** the (x-2) is under the root and then the -2 is outside

    now my textbook says you have to multiply the numerator and denominator by Squareroot (x-2) + 2

    i understand the we multiply the numerator and denominator by square root (x-2) .. but how come + 2 and not -2?


    Also can someone kindly explain how to get to teh final answer? with explanation in each step. thank you very much.
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  2. #2
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    Quote Originally Posted by alisheraz19 View Post
    Can someone kindly help me with this math problem for my calculus class.

    I'm sorry I do not have anything to post the proper form of the question (such as square root symbols) but i will try to do my best.

    Find lim x--> 6 [Squareroot(x-2) - 2] / (x-6) note** the (x-2) is under the root and then the -2 is outside
    \lim_{x\to 6}\frac{\sqrt{x-2}- 2}{x- 6}

    now my textbook says you have to multiply the numerator and denominator by Squareroot (x-2) + 2

    i understand the we multiply the numerator and denominator by square root (x-2) .. but how come + 2 and not -2?


    Also can someone kindly explain how to get to teh final answer? with explanation in each step. thank you very much.[/QUOTE]
    because (a- b)(a+ b)= a^2- b^2. Multiplying \sqrt{x-2)- 2 by \sqrt{x-2}+ 2 gives \left(\sqrt{x-2}\right)^2- \left(2\right)^2 and gets rid of the square root.
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