Limit of square root function.

  • Thread starter 11thHeaven
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  • #1
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I have to find the limit as x→∞ of √(x2+x)-x


I can't rearrange this into a form where I can put infinity into the expression and get a meaningful answer. I've tried taking out square roots to get √x( √(x+1)-√x ) but if I put infinity into this I just get ∞(∞-∞) which is meaningless.

Now I know that the limit approaches 0.5 (simply by plugging large numbers into my calculator) but I have no idea why this happens.
 

Answers and Replies

  • #2
938
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Remember the formula (x+y)(x-y) = x2 - y2? That might prove useful here.
 
  • #3
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Remember the formula (x+y)(x-y) = x2 - y2? That might prove useful here.

Are you suggesting [itex](\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)[/itex]? If so, how do I account for the [itex](\sqrt{x^2+x}+x)[/itex] factor?

Thanks for helping :)
 
  • #4
HallsofIvy
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Have you considered "multiplying and dividing by the same thing"?
 
  • #5
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Multiply and divide by √(x^2+x)+x
 
  • #6
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Multiply and divide by √(x^2+x)+x

If I do that, then I end up with [itex]\dfrac{x}{\sqrt{x^2+x}+x}[/itex], and I'm not sure how to progress from here. I can use L'Hopital's rule but then I get a horrible mess which doesn't seem to lead to any sort of answer.
 
  • #7
ehild
Homework Helper
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Divide both the numerator and denominator by x.

ehild
 
  • #8
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Divide both the numerator and denominator by x.

ehild

Ahhh. That makes the whole thing work, thanks :)
 
  • #9
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Also, could anyone explain how the limit of [itex](1+\frac{1}{x})^x=e[/itex] as x goes to infinity? I first assumed it went to 1 (as 1/x goes to 0, and (1+0)^∞=1).
 
  • #10
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Please start a new thread for each new question.
 

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