Limit Problem - # 6

[Jason76]

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

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

\(\displaystyle \infty \sin(0)\)

\(\displaystyle \infty(0)\) 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

 

[Jason76]

The answer to this is \(\displaystyle 5\pi \) The problem involved the chain rule. The chain rule was not done right, in regards to trig differentiation.

 

[Deleted member 4993]

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

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

\(\displaystyle \infty \sin(0)\)

\(\displaystyle \infty(0)\) Indeterminate

\(\displaystyle \dfrac{\dfrac{5 \pi}{x}}{\dfrac{1}{x}}\) ........................... Where did this come from?!!


\(\displaystyle \dfrac{\dfrac{d}{dx} \dfrac{5 \pi}{x}}{\dfrac{d}{dx} \dfrac{1}{x}}\)

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

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

\(\displaystyle \dfrac{\cos(\dfrac{ - 5\pi}{(\infty)^{2}})}{\dfrac{- 1}{(\infty)^{2}}} = \dfrac{1}{0}\) ??? What is this

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