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Limits With Radicals.

  1. Apr 28, 2013 #1
    When calculating the limit of the function f(x) = (x^2 + 3)/ sqrt(2x^4 + 5) as x→∞, is it correct to square the top and then place the resulting polynomial under a square root (i.e. sqrt(x^2 + 3)^2)? Then you can rewrite the problem as the square root of the limit as x→∞ of the resulting function.
    So, you'll have:
    sqrt(lim x→∞ (x^2 + 3)/ (2x^4 + 5))
    Divide above and below by x^4 and solve. Thanks.
  2. jcsd
  3. Apr 28, 2013 #2


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    Hi DAPOS! :smile:

    (try using the X2 button just above the Reply box :wink:)
    (you meant (x2 + 3)2/ (2x4 + 5) :wink:)

    Yes, that's fine. :smile:
  4. Apr 29, 2013 #3
  5. Apr 29, 2013 #4


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    While for that particular limit it is correct, squaring a function to find a limit of it is generally wrong.
  6. Apr 29, 2013 #5
    What would be the correct way to approach taking the limit of a radical? One that holds up for the majority of limits.
  7. Apr 29, 2013 #6


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    but it works if the function is always non-negative?
  8. Apr 29, 2013 #7
    It works if it isn't approaching negative infinity, right?
  9. Apr 29, 2013 #8


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    What's going on is ##\lim_{x\to\infty} f(x)^2 = L^2 \implies \lim_{x\to\infty}f(x) = \pm L##, so you need to go back and check what the sign of f(x) is for sufficiently large x.
  10. Apr 29, 2013 #9


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    Rather than square the numerator and denominator, the way to go would be to factor the expression inside the radical.

    $$ \frac{x^2 + 3}{\sqrt{2x^4 + 5}} = \frac{x^2(1 + 3/x^2)}{x^2\sqrt{2 + 5/x^4}}$$
    $$ = \frac{1 + 3/x^2}{\sqrt{2 + 5/x^4}}$$

    Now take the limit as x → ∞.
  11. Apr 30, 2013 #10
    I had been taught that method but I was just curious whether the above would actually work consistently. I can see that it won't, especially with limits to - infinity. Thanks for all the help!
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