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Thread: Trig Limit

  1. #1
    Senior Member polymerase's Avatar
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    Trig Limit

    Find the value of $\displaystyle \displaystyle\lim_{x\to{0}}\frac{\sqrt{1+tan\;x}-\sqrt{1+sin\;x}}{x^3}$

    Is there an alternate way of doing this question without using l'Hosital's Rule

    or this: $\displaystyle \displaystyle\lim_{x\to{0}}\frac{\sqrt{1+tan\;x}-\sqrt{1+sin\;x}}{x^3}\;\frac{\sqrt{1+tan\;x}+\sqrt {1+sin\;x}}{\sqrt{1+tan\;x}+\sqrt{1+sin\;x}}$ and then having to go through a long simplification before getting the answer....
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\cdot\frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^3\cos x}=$
    $\displaystyle \displaystyle=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{ 4}$
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by red_dog View Post
    $\displaystyle \displaystyle\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}\cdot\frac{1}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x(1-\cos x)}{x^3\cos x}=$
    $\displaystyle \displaystyle=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{2\sin^2\frac{x}{2}}{x^2}=\frac{1}{ 4}$
    When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?
    $\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2}$

    $\displaystyle = 2 \lim_{x \to 0} \left ( \frac{sin \left ( \frac{x}{2} \right )}{x} \right ) ^2$

    Now change the variable slightly: $\displaystyle y = \frac{x}{2}$

    Soooo....
    $\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2} = 2 \lim_{y \to 0} \left ( \frac{sin(y)}{2y} \right )^2$

    $\displaystyle = \frac{2}{4} \lim_{y \to 0} \left ( \frac{sin(y)}{y} \right )^2$

    $\displaystyle = \frac{1}{2} (1)^2 = \frac{1}{2}$

    With the other 1/2 out in front of the expression we get red_dog's answer of 1/4.

    -Dan
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by topsquark View Post
    $\displaystyle \lim_{x \to 0} \frac{2 sin^2 \left ( \frac{x}{2} \right )}{x^2}$
    $\displaystyle \lim_{x \to 0} \frac{{2\sin ^2 \dfrac{x}{2}}}{{x^2 }} = \frac{1}{2}\lim_{x \to 0} \left( {\frac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2 = \frac{1}{2}.$

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  6. #6
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    Quote Originally Posted by Krizalid View Post
    He is a physicist leave him alone!
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \lim_{x \to 0} \frac{{2\sin ^2 \dfrac{x}{2}}}{{x^2 }} = \frac{1}{2}\lim_{x \to 0} \left( {\frac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^2 = \frac{1}{2}.$

    I always miss the simple way...

    -Dan
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Alternatively, stepping in here

    $\displaystyle \lim_{x\to{0}}\frac{\tan(x)-\sin(x)}{2x^3}$

    $\displaystyle \tan(x)-\sin(x)=\bigg[x+\frac{x^3}{3}+\cdots\bigg]-\bigg[x-\frac{x^3}{6}+\cdots\bigg]\sim\frac{x^3}{2}$

    giving us

    $\displaystyle \lim_{x\to{0}}\frac{\frac{x^3}{2}}{2x^3}=\frac{1}{ 4}$
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  9. #9
    GAMMA Mathematics
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    Quote Originally Posted by polymerase View Post
    When you take the limit of $\displaystyle \frac{2\sin^2\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\displaystyle \frac{1}{4}$?
    It actually equals zero over zero, but by using L'Hopital's rule, you evaluate the zero limit of:

    $\displaystyle \frac{-2\cos(\frac{x}{2})sin(\frac{x}{2})}{2x}$ which is again zero over zero, so one more time...

    $\displaystyle f(x) = \frac{2\sin(\frac{x}{2})sin(\frac{x}{2}) - 2\cos^2(\frac{x}{2})}{4}$

    $\displaystyle \lim_{x \rightarrow 0} f(x) = \frac{2}{4}$

    Multiply that by the rest of the original limit and you have one fourth.
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