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Thread: Finding the limit of sin

  1. #1
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    Unhappy Finding the limit of sin

    Hi all, I'm new to the forum but will be using it for the foreseeable future. Right now I'm running into a logic problem that I can't wrap my brain around. Here it is.

    Find the limit of f(x) as x approaches 0. The function is (sin(2x))/(8x)

    I understand this almost factors out to be simply 0/0, which is undefined, but I don't know how to use L'Hospital's rule when dealing with Sin, Cos, Tan.

    Any help would be appreciated!
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  2. #2
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    Re: Finding the limit of sin

    Note $\dfrac{\sin(2x)}{8x} = \dfrac{1}{4} \cdot \dfrac{\sin(2x)}{2x}$

    also, you should be familiar with this limit ...

    $\displaystyle \lim_{\theta \to 0} \dfrac{\sin{\theta}}{\theta} = 1$
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  3. #3
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    Re: Finding the limit of sin

    Quote Originally Posted by Tyro View Post
    Here it is.
    Find the limit of f(x) as x approaches 0. The function is (sin(2x))/(8x)
    This is one of the most impotent concepts in limits so learn it well.
    $\displaystyle\lim_{u\to 0}\dfrac{\sin(u)}{u}=1$

    So write your problem as
    $\displaystyle\lim_{x\to 0}\frac{1}{4}\left(\dfrac{\sin(2x)}{2x}\right)=?$
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    Re: Finding the limit of sin

    That is.....extremely convenient. So as long as I can make the numerator and the denominator the same, (besides the sin), it will always be 1?

    So in the case of another example, like the limit of f(x) as x approaches 0. The function is (Sin(9x))/(6x), I can simply multiply both sides by 3/2 to make the function (Sin(3x))/(3x), which is 1, then just multiply 1 by 3/2 to get 1.5?
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    Re: Finding the limit of sin

    Quote Originally Posted by Tyro View Post
    That is.....extremely convenient. So as long as I can make the numerator and the denominator the same, (besides the sin), it will always be 1?

    So in the case of another example, like the limit of f(x) as x approaches 0. The function is (Sin(9x))/(6x), I can simply multiply both sides by 3/2 to make the function (Sin(3x))/(3x), which is 1, then just multiply 1 by 3/2 to get 1.5?
    You want $\displaystyle\lim_{x\to 0}\frac{3}{2}\left(\dfrac{\sin(9x)}{9x}\right)=?$
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    Re: Finding the limit of sin

    Quote Originally Posted by Tyro View Post
    That is.....extremely convenient. So as long as I can make the numerator and the denominator the same, (besides the sin), it will always be 1?

    So in the case of another example, like the limit of f(x) as x approaches 0. The function is (Sin(9x))/(6x), I can simply multiply both sides by 3/2 to make the function (Sin(3x))/(3x), which is 1, then just multiply 1 by 3/2 to get 1.5?
    No! sin(9x) is not 3sin(3x).
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    Re: Finding the limit of sin

    How do I make it nicer? so for the example of (Sin(9x))/(6x).
    How do I solve that?

    edit: due to some horribly misconceived rule I just concocted, if x is approaching 0 and you have a fun sin limit equation with a quotient, you just take the numbers out of the quotient for your answer. It might not work with all possible combinations, but I've found it works with many. For this example, as x approaches 0, (sin(9x))/(6x), your answer would be 9/6, or simplified, 3/2.
    Last edited by Tyro; Jul 27th 2017 at 11:55 AM.
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    Re: Finding the limit of sin

    Quote Originally Posted by Tyro View Post
    How do I make it nicer? so for the example of (Sin(9x))/(6x).
    How do I solve that?
    Just like Plato showed:

    $\displaystyle \lim_{x \to 0} \dfrac{\sin (9x)}{6x} = \lim_{x \to 0} \dfrac{\sin (9x)}{6x}\dfrac{9x}{9x} = \lim_{x \to 0} \dfrac{\sin (9x)}{9x} \dfrac{9\cancel{x}}{6\cancel{x}} = \dfrac{9}{6}\lim_{x \to 0} \dfrac{\sin (9x)}{9x}$
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  9. #9
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    Re: Finding the limit of sin

    You should also know this:

    $\displaystyle \lim_{x \to 0} \dfrac{1-\cos x}{x^2} = \dfrac{1}{2}$

    This is because:
    $\begin{align*}\displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} & = \lim_{x \to 0} \dfrac{1-\cos x}{x^2}\dfrac{1+\cos x}{1+\cos x} \\ & = \lim_{x \to 0} \dfrac{1-\cos^2 x}{x^2(1+\cos x)} \\ & = \lim_{x \to 0} \dfrac{\sin^2 x}{x^2(1+\cos x)} \\ & = \left(\lim_{x \to 0} \dfrac{\sin x}{x} \right)^2 \left(\lim_{x \to 0} \dfrac{1}{1+\cos x}\right) = 1^2 \dfrac{1}{2} = \dfrac{1}{2}\end{align*}$
    Last edited by SlipEternal; Jul 27th 2017 at 12:31 PM.
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