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About LinkBacksYou did not differentiate the numerator correctly. You should review the Chain Rule.
An easier way to do the problem:
 & = \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*})
Are you familiar with the limit? It was fairly fundamental in the proof that the derivative of
is
. All I did was use that limit. Since
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:
 & = \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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