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Question Number 102987 by bobhans last updated on 12/Jul/20

$$\underset{x\to 0}{\lim }\frac{x^2\sin \left(x^{-4}\right)}{x}?$$

Answered by bemath last updated on 12/Jul/20

$$\underset{x\to 0}{\lim }\frac{2x\sin \left(x^{-4}\right)-4x^{-3}\cos \left(x^{-4}\right)}{1}$$$$thefirstlimitconvergestozeroby$$$$squeezetest,butthesecondlimit$$$$isdivergent(because{x}^{-3}goesto$$$$infinityasx\to 0and\cos \left(x^{-4}\right)does$$$$notgotozero.$$$$\underset{x\to 0}{\lim }\frac{2x\sin \left(x^{-4}\right)-4x^{-3}\cos \left(x^{-4}\right)}{1}$$$$diverges$$

Commented by Dwaipayan Shikari last updated on 12/Jul/20

$$But\underset{x\to 0}{\lim }\frac{x^2}{x}\left(sin{x}^{-4}\right)=x\times sin\left(x^{-4}\right)=0$$$$assin\left(x^{-4}\right)isafinitenumber$$

Commented by bramlex last updated on 12/Jul/20

$$\mathrm{this}\mathrm{is}\mathrm{explanation}\mathrm{why}$$$${\mathrm{L}}^{\prime }\mathrm{Hopital}{\mathrm{rule}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{work}\mathrm{to}\mathrm{this}$$$$\mathrm{equation}⟨\mathrm{this}\mathrm{is}\mathrm{Real}\mathrm{analysis}$$$$\mid \stackrel{\bullet }{⌣}\mid \stackrel{\bullet }{⌣}\mid ⟩$$

Commented by abdomsup last updated on 12/Jul/20

$$hospitalruledontworkinthiscase$$$$becauseheistired....!$$

Answered by mathmax by abdo last updated on 12/Jul/20

$$\mathrm{we}\mathrm{have}\mathrm{for}\mathrm{x}\ne 0\frac{{\mathrm{x}}^2\sin \left(\frac{1}{{\mathrm{x}}^4}\right)}{\mathrm{x}}=\mathrm{x}\sin \left(\frac{1}{{\mathrm{x}}^4}\right)\mathrm{and}\mid \mathrm{xsin}\left(\frac{1}{{\mathrm{x}}^4}\right)\mid ⩽\mid \mathrm{x}\mid \Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{x}}^2\sin ({\mathrm{x}}^{-4)}}{\mathrm{x}}=0$$

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