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Math Help - Help evaluating limits

  1. #1
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    Help evaluating limits

    I'm having difficulty evaluating the following limits:

    lim x->4

    2-√x
    _______
    3-√2x+1

    and

    lim x->0

    2^2x - 2^x
    ___________
    2^x-1

    I'd really appreciate any help with these. I know I need to multiply by the conjugate radical in the first one, but after that I'm stuck. With the second, I'm really not sure what to do. Thanks in advance
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  2. #2
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    \lim_{x\rightarrow{0}}\frac{2^{2x}-2^{x}}{2^{x}-1}

    I'd really appreciate any help with these. I know I need to multiply by the conjugate radical in the first one, but after that I'm stuck. With the second, I'm really not sure what to do. Thanks in advance
    Factor out 2^x: \frac{2^{x}(2^{x}-1)}{2^{x}-1}=2^{x}

    This simplifies down to \lim_{x\rightarrow{0}}2^{x}

    Now, the limits easy?.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Factor out 2^x: \frac{2^{x}(2^{x}-1)}{2^{x}-1}=2^{x}

    This simplifies down to \lim_{x\rightarrow{0}}2^{x}

    Now, the limits easy?.
    That's what I thought... but wouldn't that make the limit equal to 1, because 2^0 = 1? The back of my textbook has the answer as 16. Did I do something wrong?
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  4. #4
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    No, you're correct. The limit is 1. If that is the problem, then the book is wrong.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    I'm having difficulty evaluating the following limits:

    lim x->4

    2-√x
    _______
    3-√2x+1
    you really should use brackets after the square root symbol so we know everything that is being rooted. if you look carefully, your problem looks like \frac {2 - \sqrt {x}}{3 - \sqrt {2x} + 1} when i know it's supposed to be \frac {2 - \sqrt {x}}{3 - \sqrt {2x + 1}}. in this case it is easy to see what the problem should be, but that won't always be the case. use parenthesis to clarify, it will save you and the person responding to you a lot of frustration.

    Hint. Rationalize the denominator. some algebraic manipulation would be needed after this. Think: "difference of two squares"
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    you really should use brackets after the square root symbol so we know everything that is being rooted. if you look carefully, your problem looks like \frac {2 - \sqrt {x}}{3 - \sqrt {2x} + 1} when i know it's supposed to be \frac {2 - \sqrt {x}}{3 - \sqrt {2x + 1}}. in this case it is easy to see what the problem should be, but that won't always be the case. use parenthesis to clarify, it will save you and the person responding to you a lot of frustration.

    Hint. Rationalize the denominator. some algebraic manipulation would be needed after this. Think: "difference of two squares"

    I'm sorry, I know it looks confusing. I'll use brackets from now on. I'm hoping as soon as I find a few minutes where I'm not at school or doing homework, I can learn how to insert equations/type math nicely so that at least when I come on here to ask inane questions it's in proper format

    I rationalized to:

    6 + 2(√2x+1) - 3(√x) - (√x)(√2x+1)
    ______________________________

    8-2x

    My issue is that I don't know how to multiply those last two terms: (√x)( √2x+1). Does the sign cancel out?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    I'm sorry, I know it looks confusing. I'll use brackets from now on. I'm hoping as soon as I find a few minutes where I'm not at school or doing homework, I can learn how to insert equations/type math nicely so that at least when I come on here to ask inane questions it's in proper format

    I rationalized to:

    6 + 2(√2x+1) - 3(√x) - (√x)(√2x+1)
    ______________________________

    8-2x

    My issue is that I don't know how to multiply those last two terms: (√x)( √2x+1). Does the sign cancel out?
    remember, we are concerned with the bottom, we worry about simplifying the top after we make sure the bottom does not go to zero. in the above formula, the bottom still goes to zero. more manipulation is necessary. don't multiply out the top, keep it in factored form, and then, think "difference of two squares"

    see here to get a copy of the LaTex tutorial and learn how to type "pretty math"
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    remember, we are concerned with the bottom, we worry about simplifying the top after we make sure the bottom does not go to zero. in the above formula, the bottom still goes to zero. more manipulation is necessary. don't multiply out the top, keep it in factored form, and then, think "difference of two squares"

    see here to get a copy of the LaTex tutorial and learn how to type "pretty math"
    Thank you! I finally got it. I completely overlooked the fact that 4-1x is a difference of squares. And thanks for the link as well
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by starswept View Post
    Thank you! I finally got it. I completely overlooked the fact that 4-1x is a difference of squares. And thanks for the link as well
    you're most welcome. glad i could be of assistance. good luck in your class
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