lim-x-0-0-x-2-4-t-3-dt-x-2-

Question Number 78526 by TawaTawa last updated on 18/Jan/20

Commented by mr W last updated on 18/Jan/20

lim_(x→0) [((∫_( 0) ^( x^2 ) (√(4 + t^3 )) dt)/x^2 )] =lim_(x→0) [(( (√(4 +(x^2 )^3 )) 2x)/(2x))] =lim_(x→0) [(√(4+x^6 ))] =(√4) =2

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\int_{\:\:\mathrm{0}} ^{\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\sqrt{\mathrm{4}\:+\:\boldsymbol{\mathrm{t}}^{\mathrm{3}} }\:\:\boldsymbol{\mathrm{dt}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\:\sqrt{\mathrm{4}\:+\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\mathrm{2}{x}}{\mathrm{2}{x}}\right] \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\sqrt{\mathrm{4}+{x}^{\mathrm{6}} }\right] \\ $$$$=\sqrt{\mathrm{4}} \\ $$$$=\mathrm{2} \\ $$

Commented by TawaTawa last updated on 18/Jan/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 18/Jan/20

what about lim_(x→0) [((∫_( x) ^( x^2 ) (√(4 + t^3 )) dt)/x^2 )] ?

$${what}\:{about} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\int_{\:\:{x}} ^{\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\sqrt{\mathrm{4}\:+\:\boldsymbol{\mathrm{t}}^{\mathrm{3}} }\:\:\boldsymbol{\mathrm{dt}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right]\:? \\ $$

Commented by TawaTawa last updated on 18/Jan/20

I was about to ask a question sir. did you use the integration sign. how come you get (√(4 + (x^2 )^3 )) × 2x

$$\mathrm{I}\:\mathrm{was}\:\mathrm{about}\:\mathrm{to}\:\mathrm{ask}\:\mathrm{a}\:\mathrm{question}\:\mathrm{sir}. \\ $$$$\mathrm{did}\:\mathrm{you}\:\mathrm{use}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{sign}.\:\:\mathrm{how}\:\mathrm{come}\:\mathrm{you}\:\mathrm{get}\:\:\:\sqrt{\mathrm{4}\:+\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:×\:\mathrm{2x} \\ $$

Commented by TawaTawa last updated on 18/Jan/20

I will solve your new question sir, if i understand how you integrate

$$\mathrm{I}\:\mathrm{will}\:\mathrm{solve}\:\mathrm{your}\:\mathrm{new}\:\mathrm{question}\:\mathrm{sir},\:\mathrm{if}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{how} \\ $$$$\mathrm{you}\:\mathrm{integrate} \\ $$

Commented by mr W last updated on 18/Jan/20

$${see}\:{Q}#\mathrm{78021} \\ $$

Commented by john santu last updated on 18/Jan/20

lim_(x→0) (((√(4+x^6 )) ×2x−(√(4+x^3 )) ×1)/(2x)) = lim_(x→0) (√(4+x^6 )) −lim_(x→0) ((x(√((4/x^2 )+x)))/(2x))=−∞

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{4}+{x}^{\mathrm{6}} }\:×\mathrm{2}{x}−\sqrt{\mathrm{4}+{x}^{\mathrm{3}} }\:×\mathrm{1}}{\mathrm{2}{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\mathrm{4}+{x}^{\mathrm{6}} }\:−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\sqrt{\frac{\mathrm{4}}{{x}^{\mathrm{2}} }+{x}}}{\mathrm{2}{x}}=−\infty \\ $$

Commented by mr W last updated on 18/Jan/20

correct sir! but both −∞ and +∞, so just =∞.

$${correct}\:{sir}! \\ $$$${but}\:{both}\:−\infty\:{and}\:+\infty,\:{so}\:{just}\:=\infty. \\ $$

Commented by jagoll last updated on 18/Jan/20

$${yes}\:{sir} \\ $$

Commented by TawaTawa last updated on 18/Jan/20

Checked sir, i get the approach now. = lim_(x→0) [(((√(4 + (x^2 )^3 )) × 2x − (√(4 + x^3 )) × 1)/(2x))] {The power of x decreases} = [(((√(4 + 0)) × 2(0) − (√(4 + 0)))/(2(0)))] = [(( − (√4))/(2(0)))] = − (2/0) = ∞

$$\mathrm{Checked}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{get}\:\mathrm{the}\:\mathrm{approach}\:\mathrm{now}. \\ $$$$=\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\frac{\sqrt{\mathrm{4}\:+\:\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:×\:\mathrm{2x}\:\:\:−\:\:\sqrt{\mathrm{4}\:+\:\mathrm{x}^{\mathrm{3}} }\:×\:\mathrm{1}}{\mathrm{2x}}\right]\:\:\:\left\{\mathrm{The}\:\mathrm{power}\:\mathrm{of}\:\mathrm{x}\:\mathrm{decreases}\right\} \\ $$$$=\:\:\:\:\left[\frac{\sqrt{\mathrm{4}\:+\:\mathrm{0}}\:×\:\mathrm{2}\left(\mathrm{0}\right)\:\:−\:\:\sqrt{\mathrm{4}\:+\:\:\mathrm{0}}}{\mathrm{2}\left(\mathrm{0}\right)}\right] \\ $$$$=\:\:\:\:\left[\frac{\:\:−\:\:\sqrt{\mathrm{4}}}{\mathrm{2}\left(\mathrm{0}\right)}\right] \\ $$$$=\:\:\:\:−\:\frac{\mathrm{2}}{\mathrm{0}} \\ $$$$=\:\:\:\:\infty \\ $$

Commented by mathmax by abdo last updated on 18/Jan/20

let f(x)=(1/x^2 ) ∫_x ^x^2 (√(4+t^3 ))dt ∃c_x ∈]x,x^2 [ / f(x)=((√(4+c_x ^2 ))/x^2 ) ∫_x ^x^2 dt =(√(4+c_x ^2 ))×((x^2 −x)/x^2 ) =(√(4+c_x ^2 )) ×(1−(1/x)) ⇒lim_(x→0^+ ) f(x)=−∞

$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\int_{{x}} ^{{x}^{\mathrm{2}} } \sqrt{\mathrm{4}+{t}^{\mathrm{3}} }{dt}\:\:\:\:\: \\ $$$$\left.\exists{c}_{{x}} \:\in\right]{x},{x}^{\mathrm{2}} \left[\:/\:{f}\left({x}\right)=\frac{\sqrt{\mathrm{4}+{c}_{{x}} ^{\mathrm{2}} }}{{x}^{\mathrm{2}} }\:\int_{{x}} ^{{x}^{\mathrm{2}} } \:{dt}\:=\sqrt{\mathrm{4}+{c}_{{x}} ^{\mathrm{2}} }×\frac{{x}^{\mathrm{2}} −{x}}{{x}^{\mathrm{2}} }\right. \\ $$$$=\sqrt{\mathrm{4}+{c}_{{x}} ^{\mathrm{2}} }\:×\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right)=−\infty \\ $$

Commented by TawaTawa last updated on 18/Jan/20