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Math Help - Some Limit Questions

  1. #1
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    Some Limit Questions

    Having some issues with these questions:

    1.) limit as x ---->4 \frac{x^2-x-12}{2x^2-5x-12}
    I know I can factor the numerator to have: (x-4 (x+3).

    2.) limit as x---->infinity \frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}

    Do I find a common denominator first, subtract the fractions, and take it from there?

    3.) limit as x---->2 \frac{\sqrt{x+2}-2}{x-2}

    I think I use the conjugate here.

    Well, that is how I tacked the problem. Will post my new answers shortly if I think of a better way to answer them

    TIA!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mvho View Post
    Having some issues with these questions:

    1.) limit as x ---->4 \frac{x^2-x-12}{2x^2-5x-12}
    I know I can factor the numerator to have: (x-4 (x+3).
    The denominator factors to (2x+3)(x-4). Can you take it from here?

    2.) limit as x---->infinity \frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}

    Do I find a common denominator first, subtract the fractions, and take it from there?
    Note that if you multiply the first fraction by \frac{\frac{1}{x^2}}{\frac{1}{x^2}}, you get \frac{1}{1-\frac{1}{x^2}}; if you multiply the second fraction by \frac{\frac{1}{x^4}}{\frac{1}{x^4}}, you get \frac{\frac{1}{x}+\frac{1}{x^4}}{1+\frac{1}{x^4}}

    Thus, \lim_{x\to\infty}\frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}=\lim_{x\to\infty}\frac{1}{1-\frac{1}{x^2}}-\frac{\frac{1}{x}+\frac{1}{x^4}}{1+\frac{1}{x^4}}

    Also note that \lim_{x\to\infty}\frac{1}{x^n}=0 whenever n\geq1.

    Can you take it from here?

    3.) limit as x---->2 \frac{\sqrt{x+2}-2}{x-2}

    I think I use the conjugate here.

    Well, that is how I tacked the problem. Will post my new answers shortly if I think of a better way to answer them

    TIA!
    Yes, you would use conjugate here. How did you do with this one?


    I hope this all made sense to you...
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  3. #3
    MHF Contributor red_dog's Avatar
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    1) Factor the denominator.

    2x^2-5x-12=2x^2-8x+3x-12=2x(x-4)+3(x-4)=(x-4)(2x+3)

    2)In this case you can use the limits of the rational functions when x\to\infty:

    - if both nomerator and denominator have the same degree then the limit is \frac{a_n}{b_n}, where a_n,b_n are the dominant coefficients.

    -if the denominator has a degree greater then numerator, then the limit is 0.

    3) \lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}=\lim_{x\to 2}\frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{x-2}=\lim_{x\to 2}\frac{x-2}{(x-2)(\sqrt{x+2}+2)}
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  4. #4
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    Thanks guys! Chris, that made a lot of sense. I just suck hardcore at factoring. When I saw that question, I knew the denominator would screw me over.

    For the 2nd question, since x is going to infinity essentially a small number divided by a really big number will approach 0.

    For the 3rd one, looks like I forgot to multiply -2 x 2, but I worked it out. I got \frac{1}{4} as my answer.

    As for the 1st one, I am really not good at factoring as embarrassed to say I don't see how you factored it. Is there some kind of method?

    Like, how did you go from:

    2x^2-5x-12 to 2x^2-8x+3x-12

    From there, I can see how you factor it easily. Just can't make that initial step


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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by red_dog View Post

    3) \lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}={\color{red}{\lim_{x\to 2}\frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{x-2}}}=\lim_{x\to 2}\frac{x-2}{(x-2)(\sqrt{x+2}+2)}


    Do you mean this?

    \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2}  - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt {x + 2}  - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}}.

    or

    \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2}  - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2}  - 2}}{{x + 2 - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2}  - 2}}{{\left( {\sqrt {x + 2}  - 2} \right)\left( {\sqrt {x + 2}  + 2} \right)}}.
    Last edited by DeMath; August 1st 2009 at 03:53 AM.
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  6. #6
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    Quote Originally Posted by mvho View Post
    I am really not good at factoring as embarrassed to say I don't see how you factored it. Is there some kind of method?
    Yes.
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