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Limits - square root

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  1. Jul 15, 2016 #1

    Rectifier

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    The problem
    $$ \lim_{x \rightarrow \infty} \left( \sqrt{x+1} - \sqrt{x} \right) $$

    The attempt
    ## \left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{x+1 - x }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{x \left( \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} \right) } = \frac{1}{x} \frac{1 }{ \sqrt{\frac{1}{x}+\frac{1}{x^2}} + \sqrt{\frac{1}{x}} } \\ \rightarrow 0 \cdot \frac{1}{0 + 0} ## for ## x \rightarrow \infty ## division with 0 is undefined. Can somone help me calculate this limit.

    And no, I cant use l'Hopitals rule so please don't mention it in this thread.
     
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  3. Jul 15, 2016 #2

    Math_QED

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    ## \left( \sqrt{x+1} - \sqrt{x} \right) = \frac{\left( \sqrt{x+1} - \sqrt{x} \right)\left( \sqrt{x+1} + \sqrt{x} \right) }{\left( \sqrt{x+1} + \sqrt{x} \right) } = \frac{1 }{\left( \sqrt{x+1} + \sqrt{x} \right) } ##.

    Until here, what you do leads to the answer. When you fill in infinity right now, you don't get an undetermined form, so the answer is ...?
     
  4. Jul 15, 2016 #3

    Ray Vickson

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    Note that
    [tex] \frac{x+1-x}{\sqrt{x}+\sqrt{x+1}} = \frac{1}{\sqrt{x}+\sqrt{x+1}} [/tex]
    In the second form, what happens to the numerator when ##x \to +\infty##? What happens to the denominator?

    What dictator forbids you from using l'Hospital's rule?
     
  5. Jul 15, 2016 #4

    Mark44

    Staff: Mentor

    I haven't worked this problem, but sometimes L'Hopital's rule doesn't get you anywhere. I've seen similar problems where two applications of L'H gets you right back to where you started.
     
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