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lim-x-0-d-dx-0-x-sin-t-3-dt-2x-4-




Question Number 116019 by bemath last updated on 30/Sep/20
lim_(x→0)  (((d/dx) ∫_0 ^x  sin (t^3 ) dt)/(2x^4 )) ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\mathrm{sin}\:\left({t}^{\mathrm{3}} \right)\:{dt}}{\mathrm{2}{x}^{\mathrm{4}} }\:? \\ $$
Commented by bemath last updated on 30/Sep/20
lim_(x→0) ((sin (x^3 ))/(2x^4 )) = lim_(x→0)  ((sin (x^3 ))/x^3 ) .(1/(2x))  = ± ∞ ? (is it true ?)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)}{\mathrm{2}{x}^{\mathrm{4}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }\:.\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$=\:\pm\:\infty\:?\:\left({is}\:{it}\:{true}\:?\right) \\ $$
Commented by MJS_new last updated on 30/Sep/20
lim_(x→0)  (((d/dx)[∫_0 ^x sin (t^3 ) dt])/(2x^4 )) =lim_(x→0)  ((sin x^3 )/(2x^4 )) =  =lim_(x→0)  ((3cos x^3 )/(8x)) doesn′t exist
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[\underset{\mathrm{0}} {\overset{{x}} {\int}}\mathrm{sin}\:\left({t}^{\mathrm{3}} \right)\:{dt}\right]}{\mathrm{2}{x}^{\mathrm{4}} }\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{4}} }\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3cos}\:{x}^{\mathrm{3}} }{\mathrm{8}{x}}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$
Commented by bemath last updated on 30/Sep/20
thank you sir for your correction
$${thank}\:{you}\:{sir}\:{for}\:{your}\:{correction} \\ $$

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