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Find the limit

  1. Aug 13, 2006 #1
    Find limit of f(x) when x approaches 0.

    Given that f(x) is (1/x^3){(1+tanx)^0.5 - (1+sinx)^0.5}

    The given answer is 0.25, but can somebody show me the solutions? I try the conjugate, and nothing works. Then I try to substitute t=(1+tanx)^0.5 and others, can't work, too.
  2. jcsd
  3. Aug 13, 2006 #2


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    Homework Helper

    If you multiply numerator and denominator with the complement expression, you get after simplifying:

    [tex]\frac{{\frac{{\tan x - \sin x}}{{x^3 }}}}{{\sqrt {1 + \tan x} + \sqrt {1 + \sin x} }}[/tex]

    Replacing sin(x) and tan(x) by the first terms of their Taylor series arround 0 so that the difference isn't 0, is x³/2. So you get:

    \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\tan x - \sin x}}{{x^3 }}}}{{\sqrt {1 + \tan x} + \sqrt {1 + \sin x} }} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{2}}}{{\sqrt {1 + \tan x} + \sqrt {1 + \sin x} }}
  4. Aug 13, 2006 #3


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    Science Advisor

    Wait a minute, what is your f(x)?
    Is it
    [tex]\frac {\sqrt{1+tanx} - \sqrt{1+sinx}}{x^3}[/tex]?
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