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Math Help - Help needed in Limits

  1. #1
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    Help needed in Limits

    Please, kindly explain the solution step by step.

    Limit problems are in \frac{0}{0} form.

    1. \lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}

    2. \lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}

    3. \lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}
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  2. #2
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    Hi

    Quote Originally Posted by MathematicalDick View Post
    1. \lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}
    The fact that you find 0/0 means that you can factor out (x+2) to both numerator and denominator
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  3. #3
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    Quote Originally Posted by MathematicalDick View Post
    2. \lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}
    http://www.mathhelpforum.com/math-he...tml#post480676
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  4. #4
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    Quote Originally Posted by MathematicalDick View Post
    Please, kindly explain the solution step by step.

    Limit problems are in \frac{0}{0} form.

    1. \lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}

    2. \lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}

    3. \lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}
    2. \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} = \left(\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}}\right)\left(\frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}}\right)

     = \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{x^2 - a^2}

     = \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}

     = \left[\frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}\right]\left(\frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}\right)

     = \frac{(x - a)\sqrt{x^2 - a^2}}{(x - a)(x + a)(\sqrt{x} + \sqrt{a})}

     = \frac{\sqrt{x^2 - a^2}}{(x + a)(\sqrt{x} + \sqrt{a})}.

    Now make x \to a.
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  5. #5
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    Quote Originally Posted by MathematicalDick View Post
    Please, kindly explain the solution step by step.

    Limit problems are in \frac{0}{0} form.

    1. \lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}

    2. \lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}

    3. \lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}
    3. \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4} = \frac{(x - \sqrt{2})(x + \sqrt{2})}{(x - \sqrt{2})(x + 2\sqrt{2})}

     = \frac{x + \sqrt{2}}{x + 2\sqrt{2}}.

    Now make x \to \sqrt{2}.
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  6. #6
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    I absolutely forgot factorization. I factored x^3-3x+2 as (x-1)(x^2+x-2) before you gave the the hint that both the Nr and the Dr will be factored by x+2.

    I finally got the answer \lim_{x\to \-2} \frac{(x+2)(x^2-x+6)}{(x+2)(x^2-2x+1)} = \frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1} = \frac{4}{3}


    I want to revise the factorization, may I request you to kindly send me the tutorial link of these kind of factorials in this community. Thanks
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  7. #7
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    Quote Originally Posted by Prove It View Post
    2. \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 -  a^2}} = \left(\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 -  a^2}}\right)\left(\frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 -  a^2}}\right)

     = \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{x^2 - a^2}

     = \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x +  a)}

     = \left[\frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x +  a)}\right]\left(\frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} +  \sqrt{a}}\right)

     = \frac{(x - a)\sqrt{x^2 - a^2}}{(x - a)(x + a)(\sqrt{x} +  \sqrt{a})}

     = \frac{\sqrt{x^2 - a^2}}{(x + a)(\sqrt{x} + \sqrt{a})}.

    Now make x \to a.
    hmmm

    Why so long? How about this?

     \lim_{x\to \ a} \left(\frac  {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}\right) X \left(\frac  {\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\right)

    = \lim_{x\to \ a}\frac{x-a}{(\sqrt{x}+\sqrt{a})\sqrt{x^2-a^2}}

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

    Now make x \to a
    Last edited by MathematicalDick; March 27th 2010 at 03:07 AM.
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  8. #8
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    hi

    this is long method










    .

    Now make .

    this is the short cut method


    X

    =

    =

    = ....its the easy way to answer your question...










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  9. #9
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    Thanks. But I have already got the answer and posted the same thing just above your post.
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