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Math Help - Evaluate the limit.

  1. #1
    Super Member General's Avatar
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    Evaluate the limit.

    Hi
    Without using L`Hospital`s Rule, Evaluate the following limit:
    \lim_{x \rightarrow 0} \frac{ \sqrt{1+tanx} - \sqrt{1+sinx} }{x^3}
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  2. #2
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    I posted the solution in parts. See if you can continue by yourself after every step you see.


    <br />
\lim_{x\rightarrow 0} \frac{\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}}}{x^3} = <br />

    <br />
= \lim_{x\rightarrow 0} \frac{(\sqrt{1+\tan{x}}-\sqrt{1+\sin{x}})(\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}  })}{x^3 (\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}})} =<br />

    Spoiler:

    <br />
=\lim_{x\rightarrow 0}\frac{1}{\sqrt{1+\tan{x}}+\sqrt{1+\sin{x}}} \lim_{x\rightarrow 0} \frac{\tan{x}-\sin{x}}{x^3} =<br />


    Spoiler:

    <br />
= \frac{1}{2}\lim_{x\rightarrow 0} \frac{1}{\cos{x}} \lim_{x\rightarrow 0} \frac{\sin{x}(1-\cos{x})}{x^3} = \frac{1}{2} \lim_{x\rightarrow 0} \frac{\sin{x}(2\sin^2{\frac{x}{2}})}{x^3} =<br />


    Spoiler:

    <br />
= 2\lim_{x\rightarrow 0} \frac{\sin^3{\frac{x}{2}}\cos{\frac{x}{2}}}{x^3} = 2\lim_{x\rightarrow 0} \cos{\frac{x}{2}}\lim_{x\rightarrow 0} \frac{\sin^3{\frac{x}{2}}}{x^3} = <br />


    Spoiler:

    <br />
= \frac{1}{4}\lim_{x\rightarrow 0} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} \frac{\sin{\frac{x}{2}}}{\frac{x}{2}} = <br />


    Spoiler:

    =\frac{1}{4}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by General View Post
    Hi
    Without using L`Hospital`s Rule, Evaluate the following limit:
    \lim_{x \rightarrow 0} \frac{ \sqrt{1+tanx} - \sqrt{1+sinx} }{x^3}
    Let L denote the above limit. Clearly 2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3} (multiply top and bottom by conjugate and seprate limits). So then 2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\lim_{x\to 0}\frac{x+\tfrac{x^3}{3}+\cdots-\left(x-\tfrac{x^3}{6}+\cdots\right)}{x^3}=\lim_{x\to0}\fr  ac{\tfrac{x^3}{2}+\cdots}{x^3}=\frac{1}{2} so that L=\frac{1}{4}
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Let L denote the above limit. Clearly 2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3} (multiply top and bottom by conjugate and seprate limits). So then 2L=\lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3}=\lim_{x\to 0}\frac{x+\tfrac{x^3}{3}+\cdots-\left(x-\tfrac{x^3}{6}+\cdots\right)}{x^3}=\lim_{x\to0}\fr  ac{\tfrac{x^3}{2}+\cdots}{x^3}=\frac{1}{2} so that L=\frac{1}{4}
    Its 2 or \sqrt 2 ?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by General View Post
    Its 2 or \sqrt 2 ?
    I misread the question, and made a typo. It should be two. Although, I would suggest you look at Unbeatable's solution. He didn't take the easy way out and use Maclaurin.
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  6. #6
    Super Member General's Avatar
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    What is the general formula of Mac. series for tanx?
    i.e: the formula contains Sigma
    I know it for arctan x
    not tan x
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by General View Post
    What is the general formula of Mac. series for tanx?
    i.e: the formula contains Sigma
    I know it for arctan x
    not tan x
    It's crazy and involves Bernoulli numbers. It's not worth remembering, but just do the calculation three times (the second one vanishes) and remmeber that.
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