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Math Help - Limits to infinity

  1. #1
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    Limits to infinity

    \lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}\\\\\\ \lim_{x \to \infty} \frac{\frac{2x}{x}}{\sqrt{\frac{x}{x}+\frac{2}{x}} + \sqrt{\frac{x}{x}}}\\\\\\ \lim_{x \to \infty} \frac{2}{\sqrt{1 + \frac{2}{x}} + \sqrt{1}}\\\\\\ \lim_{x \to \infty} \frac{2}{\sqrt{1} + \sqrt{1}}\\\\\\ \lim_{x \to \infty} \frac{2}{1 + 1} = \frac{2}{2} = 1\\\\\\

    I don't understand where I did my work wrong. I divided all the terms by a common multiple to simplify them, so I don't see anywhere I could have done it incorrectly.
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  2. #2
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    Re: Limits to infinity

    In your first step, when you bring 1/x under the radicals, it needs to be 1/x^2.
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  3. #3
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    Re: Limits to infinity

    Quote Originally Posted by PhizKid View Post
    \lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}\\\\\\ \lim_{x \to \infty} \frac{\frac{2x}{x}}{\sqrt{\frac{x}{x}+\frac{2}{x}} + \sqrt{\frac{x}{x}}}
    That's where you made a mistake.

    Doing it your way would be doing it like this:

    \lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}

    = \lim_{x \to \infty} \frac{\frac{2x}{x}}{\left( \frac{1}{x} \right) \left(\sqrt{x + 2} + \sqrt{x}\right)}

    = \lim_{x \to \infty} \frac{2}{ \left \sqrt{\frac{1}{x^2}} \right) \left(\sqrt{x + 2} + \sqrt{x}\right)}

    = \lim_{x \to \infty} \frac{2}{ \left( \sqrt{\frac{1}{x^2}} \right) \left(\sqrt{x + 2} \right) + \left( \sqrt{\frac{1}{x^2}} \right) \left( \sqrt{x} \right)}

    = \lim_{x \to \infty} \frac{2}{\sqrt{\frac{1}{x^2}(x + 2)} + \sqrt{\frac{1}{x^2}(x)}}

    = \lim_{x \to \infty} \frac{2}{\sqrt{\frac{x+2}{x^2}} + \sqrt{ \frac{x}{x^2} } }

    = \lim_{x \to \infty} \frac{2}{\sqrt{\frac{x}{x^2} + \frac{2}{x^2}} + \sqrt{\frac{1}{x} } }

    = \lim_{x \to \infty} \frac{2}{\sqrt{ \frac{1}{x} + \frac{2}{x^2} } + \sqrt{ \frac{1}{x} } }.

    So it's like 2/0, which goes to positive infinity.

    Another way (a better way I think) is to divide numerator and denominator by \sqrt{x} to kill the "to infinity" happening in the denominator.

    \lim_{x \to \infty} \frac{2x}{\sqrt{x+2} + \sqrt{x}}

    =\lim_{x \to \infty} \left( \frac{2x}{\sqrt{x+2} + \sqrt{x}} \right) \left( \frac{ \frac{1}{\sqrt{x}} }{ \frac{1}{\sqrt{x}} } \right)

    =\lim_{x \to \infty} \left( \frac{2x \left(\frac{1}{\sqrt{x}} \right) }{ (\sqrt{x+2} + \sqrt{x})\left(\frac{1}{\sqrt{x}} \right) } \right)

    =\lim_{x \to \infty} \frac{ \frac{2x}{\sqrt{x}} }{ \frac{\sqrt{x+2}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{x}} }

    =\lim_{x \to \infty} \frac{ 2\sqrt{x} }{ \sqrt{ \frac{x+2}{x} } + 1 }

    =\lim_{x \to \infty} \frac{ 2\sqrt{x} }{ \sqrt{ 1 + \frac{2}{x} } + 1 }.

    \text{So the numerator goes to infinity, the denominator goes to 2, so the limit of the quotient is positive infinity.}
    Last edited by johnsomeone; October 27th 2012 at 03:31 AM.
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  4. #4
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    Re: Limits to infinity

    Oh, I'm still a little confused. Our professor just told us to divide out each term by a common factor or factor some common factor out of each term, so that's just what I did. I didn't think it mattered what kind of term it was but apparently it's something very different.

    So you multiply every term by (1/x) every time?

    In your first method, how did your 1/x become sqrt(1/x^2)?
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  5. #5
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    Re: Limits to infinity

    Quote Originally Posted by PhizKid View Post
    Oh,
    So you multiply every term by (1/x) every time?
    In your first method, how did your 1/x become sqrt(1/x^2)?
    If x>0 then \frac{1}{x}=\frac{1}{\sqrt{x^2}}
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  6. #6
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    Re: Limits to infinity

    another approach:

    we may assume x > 2, since we are taking the limit at "positive infinity".

    thus:

    \sqrt{x} < \sqrt{x+2}, and 2x = x + x > x + 2

    so:

    \sqrt{x+2} + \sqrt{x} < 2\sqrt{x+2} and:

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

    since:

    \lim_{x \to \infty} \frac{\sqrt{x+2}}{2} = \infty

    \lim_{x \to \infty} \frac{2x}{\sqrt{2+x}+\sqrt{x}} = \infty (it's always (for x > 2) bigger than half the square root of x+2, right?)
    Last edited by Deveno; October 27th 2012 at 06:12 AM.
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