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Math Help - Evaluate Limit

  1. #1
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    Evaluate Limit

    Hello,
    Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals \frac{1}{4};

    \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Robb View Post
    Hello,
    Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals \frac{1}{4};

    \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}
    do you want me to give you a hint or solve it
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  3. #3
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    Solving is fine... i tried multiplying out by \frac{\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}}}{\sqrt{1+\  tan{x}}+\sqrt{1+\sin{x}}} but couldn't find anything useful from it?
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  4. #4
    MHF Contributor Amer's Avatar
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    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}

    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)

    lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}

    lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{2x^3}

    lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{2x^3}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{2x^3(cosx)}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{2x^3(1+cosx)}

    lim_{x\rightarrow 0}\frac{(sin^3x)}{4x^3}=\frac{1}{4} since lim_{x\rightarrow 0} \frac{sinx}{x} =1

    Edited :- fix a mistake in the last line thanks to "mr fantastic"
    Last edited by Amer; June 12th 2009 at 06:16 AM.
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  5. #5
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    Quote Originally Posted by Robb View Post
    Hello,
    Doing some revision, and I cant figure out what to do with this function in order to calculate the limit... I know it equals \frac{1}{4};

    \lim_{x\to0}\frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3}
    First multiply the numerator and denominator by the conjugate surd  \sqrt{1 + \tan x} + \sqrt{1 + \sin x} and simplify:

    Spoiler:

    \frac{\sin x (1 - \cos x)}{x^3 \cos x \left( \sqrt{1 + \tan x} + \sqrt{1 + \sin x} \right)}


    Then multiply the numerator and denominator by 1 + \cos x and simplify:

    Spoiler:

    \frac{\sin^3 x}{x^3 \cos x \left( \sqrt{1 + \tan x} + \sqrt{1 + \sin x} \right)}


    The well known limit \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 shuold then be applied.
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  6. #6
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    Thanks allot for your help, I didn't think to expand the \tan (x) out!
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  7. #7
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    Quote Originally Posted by Amer View Post
    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}

    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)

    lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}

    lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{2x^3}

    lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{2x^3}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{2x^3(cosx)}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{2x^3(1+cosx)}

    lim_{x\rightarrow 0}\frac{sinx(sin^3x)}{4x^3}=\frac{1}{4} since lim_{x\rightarrow 0} \frac{sinx}{x} =1

    [tex]
    There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.
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  8. #8
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by mr fantastic View Post
    There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.
    I think I should not sub zero in the limit before I finish the problem like this

    Quote Originally Posted by Amer View Post
    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}

    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)

    lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}

    lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{(\sqrt{tanx+1}+\sqrt{sinx+1})x^3}

    lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{x^3(cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})}

    lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sq  rt{sinx+1})}

    lim_{x\rightarrow 0}\frac{sinx(sin^3x)}{x^3cos(x)(1+cosx)(\sqrt{tanx  +1}+\sqrt{sinx+1})}=\frac{1}{4} since lim_{x\rightarrow 0} \frac{sinx}{x} =1

    [tex]
    I correct them or not ??
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  9. #9
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    Quote Originally Posted by Amer View Post
    Quote Originally Posted by mr fantastic View Post
    There are some minor corrections that need to be made in this solution. Rather than point them out, it will be instructive for the interested reader to make them him/herself.
    I think I should not sub zero in the limit before I finish the problem like this
    Quote Originally Posted by Amer View Post
    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3}
    lim_{x\rightarrow 0} \frac{\sqrt{tanx+1}-\sqrt{sinx+1}}{x^3} \left(\frac{\sqrt{tanx+1} + \sqrt{sinx+1}}{\sqrt{tanx+1} + \sqrt{sinx+1}}\right)
    lim_{x\rightarrow 0} \frac{tanx+1-1-sinx }{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}
    lim_{x\rightarrow 0}\frac{\dfrac{sinx}{cosx} - sinx }{(\sqrt{tanx+1}+\sqrt{sinx+1})x^3}
    lim_{x\rightarrow 0} \frac{\dfrac{sinx - sinxcosx}{cosx}}{x^3(\sqrt{tanx+1}+\sqrt{sinx+1})}
    lim_{x\rightarrow 0} \frac{sinx(1-cosx)}{x^3(cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})}
    lim_{x\rightarrow 0} \frac{sinx(1-cosx)(1+cosx)}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sq  rt{sinx+1})}
    lim_{x\rightarrow 0}\frac{{\color{red}sin^3 x}}{x^3cos(x)(1+cosx)(\sqrt{tanx+1}+\sqrt{sinx+1})  }=\frac{1}{4} since lim_{x\rightarrow 0} \frac{sinx}{x} =1
    [tex]
    I correct them or not ??
    Se the red numerator of the last line
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