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Question Number 42808 by maxmathsup by imad last updated on 02/Sep/18
let f(x) = ∫_x ^(2x) ((1+cos(t))/( (√(t^4 −t^2 +4))))dt 1) find D_f 2) calculate f^′ (x) 3) find lim_(x→+∞) f(x)
$${let}\:{f}\left({x}\right)\:=\:\int_{{x}} ^{\mathrm{2}{x}} \:\:\frac{\mathrm{1}+{cos}\left({t}\right)}{\:\sqrt{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} \:+\mathrm{4}}}{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18
f′(x)=∫_x ^(2x) (∂/∂x)(((1+cost)/( (√(t^4 −t^2 +4)))))dx+((1+cos2x)/( (√((2x)^4 −(2x)^2 +4))))×((d(2x))/dx)−((1+cosx)/( (√(x^4 −x^2 +4))))×((d(x))/dx) =2×((1+cos2x)/( (√(16x^4 −4x^2 +4))))−((1+cosx)/( (√(x^4 −x^2 +4)))) for any value of x the value 1≥cosx≥−1 and 1≥cos2x≥−1 so lim_(x→∞) ((2+2cos2x)/( (√(16x^4 −4x^2 +4))))−((1+cosx)/( (√(x^4 −x^2 +4)))) lim_(x→∞) ((p(x))/(q(x)))−((g(x))/(h(x))) when x→∞ p(x)→4 or 0 but q(x)→∞ when x→∞ g(x)→2 or 0 but h(x)→∞ solim_(x→∞) ((p(x))/(q(x)))−((g(x))/(h(x)))→0
$${f}'\left({x}\right)=\int_{{x}} ^{\mathrm{2}{x}} \frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}+{cost}}{\:\sqrt{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{4}}}\right){dx}+\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\:\sqrt{\left(\mathrm{2}{x}\right)^{\mathrm{4}} −\left(\mathrm{2}{x}\right)^{\mathrm{2}} +\mathrm{4}}}×\frac{{d}\left(\mathrm{2}{x}\right)}{{dx}}−\frac{\mathrm{1}+{cosx}}{\:\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{4}}}×\frac{{d}\left({x}\right)}{{dx}} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\:\sqrt{\mathrm{16}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}}−\frac{\mathrm{1}+{cosx}}{\:\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$${for}\:{any}\:{value}\:{of}\:{x}\:{the}\:{value}\:\mathrm{1}\geqslant{cosx}\geqslant−\mathrm{1} \\ $$$${and}\:\mathrm{1}\geqslant{cos}\mathrm{2}{x}\geqslant−\mathrm{1} \\ $$$${so}\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}{x}}{\:\sqrt{\mathrm{16}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}}−\frac{\mathrm{1}+{cosx}}{\:\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{p}\left({x}\right)}{{q}\left({x}\right)}−\frac{{g}\left({x}\right)}{{h}\left({x}\right)} \\ $$$${when}\:{x}\rightarrow\infty\:\:{p}\left({x}\right)\rightarrow\mathrm{4}\:{or}\:\mathrm{0}\:\:{but}\:{q}\left({x}\right)\rightarrow\infty \\ $$$${when}\:{x}\rightarrow\infty\:\:{g}\left({x}\right)\rightarrow\mathrm{2}\:\:{or}\:\mathrm{0}\:\:{but}\:{h}\left({x}\right)\rightarrow\infty \\ $$$${so}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{p}\left({x}\right)}{{q}\left({x}\right)}−\frac{{g}\left({x}\right)}{{h}\left({x}\right)}\rightarrow\mathrm{0} \\ $$$$ \\ $$

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