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

lim_(x→0) ((x^2 sin (x^(−4) ))/x) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} \:\mathrm{sin}\:\left({x}^{−\mathrm{4}} \right)}{{x}}\:?\: \\ $$

Answered by bemath last updated on 12/Jul/20

lim_(x→0) ((2x sin (x^(−4) )−4x^(−3) cos (x^(−4) ))/1) the first limit converges to zero by squeeze test , but the second limit is divergent (because x^(−3) goes to infinity as x→0 and cos (x^(−4) ) does not go to zero. lim_(x→0) ((2x sin (x^(−4) )−4x^(−3) cos (x^(−4) ))/1) diverges

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}\:\mathrm{sin}\:\left({x}^{−\mathrm{4}} \right)−\mathrm{4}{x}^{−\mathrm{3}} \:\mathrm{cos}\:\left({x}^{−\mathrm{4}} \right)}{\mathrm{1}} \\ $$$${the}\:{first}\:{limit}\:{converges}\:{to}\:{zero}\:{by} \\ $$$${squeeze}\:{test}\:,\:{but}\:{the}\:{second}\:{limit} \\ $$$${is}\:{divergent}\:\left({because}\:{x}^{−\mathrm{3}} \:{goes}\:{to}\right. \\ $$$${infinity}\:{as}\:{x}\rightarrow\mathrm{0}\:{and}\:\mathrm{cos}\:\left({x}^{−\mathrm{4}} \right)\:{does} \\ $$$${not}\:{go}\:{to}\:{zero}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}\:\mathrm{sin}\:\left({x}^{−\mathrm{4}} \right)−\mathrm{4}{x}^{−\mathrm{3}} \:\mathrm{cos}\:\left({x}^{−\mathrm{4}} \right)}{\mathrm{1}} \\ $$$${diverges} \\ $$

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

But lim_(x→0) (x^2 /x)(sin x^(−4) )=x×sin(x^(−4) )=0 as sin(x^(−4) ) is a finite number

$${But}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{x}}\left({sin}\:{x}^{−\mathrm{4}} \right)={x}×{sin}\left({x}^{−\mathrm{4}} \right)=\mathrm{0} \\ $$$${as}\:{sin}\left({x}^{−\mathrm{4}} \right)\:{is}\:{a}\:{finite}\:{number} \\ $$

Commented by bramlex last updated on 12/Jul/20

this is explanation why L′Hopital rule′s not work to this equation ⟨this is Real analysis ∣⌣^• ∣⌣^• ∣ ⟩

$$\mathrm{this}\:\mathrm{is}\:\mathrm{explanation}\:\mathrm{why} \\ $$$$\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule}'\mathrm{s}\:\mathrm{not}\:\mathrm{work}\:\mathrm{to}\:\mathrm{this} \\ $$$$\mathrm{equation}\:\langle\mathrm{this}\:\mathrm{is}\:\mathrm{Real}\:\mathrm{analysis} \\ $$$$\mid\overset{\bullet} {\smile}\mid\overset{\bullet} {\smile}\mid\:\rangle\: \\ $$

Commented by abdomsup last updated on 12/Jul/20

hospital rule dont work in this case because he is tired....!

$${hospital}\:{rule}\:{dont}\:{work}\:{in}\:{this}\:{case} \\ $$$${because}\:{he}\:{is}\:{tired}....! \\ $$

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

we have for x≠0 ((x^2 sin((1/x^4 )))/x) =x sin((1/x^4 )) and ∣xsin((1/x^4 ))∣≤∣x∣ ⇒ lim_(x→0) ((x^2 sin(x^(−4)) )/x) =0

$$\mathrm{we}\:\mathrm{have}\:\:\mathrm{for}\:\mathrm{x}\neq\mathrm{0}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)}{\mathrm{x}}\:=\mathrm{x}\:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)\:\:\mathrm{and}\:\mid\mathrm{xsin}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right)\mid\leqslant\mid\mathrm{x}\mid\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{x}^{\left.−\mathrm{4}\right)} \right.}{\mathrm{x}}\:=\mathrm{0} \\ $$

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