Thread: 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

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

An easier way to do the problem:

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

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.

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:

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