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Limit question

  1. Jun 11, 2007 #1
    I have been reading through Schaum's and came across a limits question that i cannot figure out. I thought it had no limit but the answer is 1/(2*sqrt(3)). I have no idea of how to use latex so I will attempt it using basic characters; hopefully not to obscure.

    lim(x>0) ((sqrt(x+3))-(sqrt(3))/x

    I cannot see how to simplify in order to get rid of the denominator x.
  2. jcsd
  3. Jun 11, 2007 #2


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    Multiply the whole thing by (sqrt(x+3)+sqrt(3))/(sqrt(x+3)+sqrt(3)). This is called multiplying by the conjugate and its often useful. The effect is to "switch" the square roots from denom to num. (or vice versa) and to take the arguments of the square roots "out of them". Understandable?
  4. Jun 11, 2007 #3
    Merci beaucoup, all is clear. If not too difficult can you give me a definition of the conjugate? Maybe its in this book...
  5. Jun 11, 2007 #4


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    "conjugate" may mean several things. But in this particular case, say, we have the expression: [tex]\sqrt{a} + \sqrt{b}[/tex], then its conjugate is: [tex]\sqrt{a} - \sqrt{b}[/tex], or [tex]\sqrt{b} - \sqrt{a}[/tex].

    This is a common practice in taking the limit of something, especially in the form 0 / 0. Since you can factor a polynomial, but not a surd. So, what you should do is to rationalize it, i.e eliminate the square roots.
  6. Jun 11, 2007 #5

    Gib Z

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    This is some circular logic here but in case you just wanted to value and not how to prove it, you could recognize that the limit you stated is the definition for the derivative of the square root function evaluated at 3.
  7. Jun 12, 2007 #6
    Thank you all. Gib Z, I have only begun limits; but I'm sure I'll get to derivatives and the like soon, only a few chapters away.