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Thread: Trig Limit Problem

  1. #1
    MHF Contributor Jason76's Avatar
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    Trig Limit Problem

    $\displaystyle x \sin(\dfrac{5\pi}{x})$

    $\displaystyle \infty \sin (\dfrac{5\pi}{\infty})$

    $\displaystyle \infty \sin(0)$

    $\displaystyle \infty(0)$ Indeterminate

    $\displaystyle \dfrac{\dfrac{5 \pi}{x}}{\dfrac{1}{x}}$


    $\displaystyle \dfrac{\dfrac{d}{dx} \dfrac{5 \pi}{x}}{\dfrac{d}{dx} \dfrac{1}{x}}$

    $\displaystyle \dfrac{\cos(\dfrac{x(0) - 5\pi(1)}{x^{2}})}{\dfrac{x(0) - (1)(1)}{x^{2}}}$

    $\displaystyle \dfrac{\cos(\dfrac{ - 5\pi}{x^{2}})}{\dfrac{- 1}{x^{2}}}$

    $\displaystyle \dfrac{\cos(\dfrac{ - 5\pi}{(\infty)^{2}})}{\dfrac{- 1}{(\infty)^{2}}} = \dfrac{1}{0}$ ??? What is this
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  2. #2
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    Re: Trig Limit Problem

    You did not differentiate the numerator correctly. You should review the Chain Rule.

    An easier way to do the problem:

    $\displaystyle \begin{align*}\lim_{x \to \infty} x\sin\left(\dfrac{5\pi}{x}\right) & = \lim_{x \to \infty} \dfrac{ \sin \left( \dfrac{5\pi}{x} \right) }{ \left( \dfrac{1}{x} \right) } \\ & = \lim_{u \to 0} \dfrac{\sin(5\pi u)}{u} \\ & = \lim_{u \to 0}5\pi \dfrac{\sin(5\pi u)}{5\pi u} \\ & = \lim_{5\pi u \to 0} 5\pi \dfrac{\sin(5\pi u)}{5\pi u} \\ & = 5\pi\end{align*}$
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  3. #3
    MHF Contributor Jason76's Avatar
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    Re: Trig Limit Problem

    The shortcut way looks interesting, but there is no change from $\displaystyle \sin$ to it's derivative of $\displaystyle \cos$ and so confusing, but thanks for the help anyways.
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  4. #4
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    Re: Trig Limit Problem

    Are you familiar with the limit $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x} = 1$? It was fairly fundamental in the proof that the derivative of $\displaystyle \sin x$ is $\displaystyle \cos x$. All I did was use that limit. Since $\displaystyle \infty$ is not a number, you cannot plug it in. If you want to use L'Hospital's Rule, here is how it would be used correctly:

    $\displaystyle \begin{align*}\lim_{x \to \infty} x\sin\left(5\pi x^{-1}\right) & = \lim_{x \to \infty} \dfrac{\sin\left(5\pi x^{-1}\right) }{x^{-1}} \\ & = \lim_{x \to \infty} \dfrac{-5\pi x^{-2}\cos\left(5\pi x^{-1}\right)}{-x^{-2}} \\ & = \lim_{x \to \infty} 5\pi \cos\left(5 \pi x^{-1}\right) \\ & = 5\pi \cos(0) \\ & = 5\pi\end{align*}$
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