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Question Number 145806 by bramlexs22 last updated on 08/Jul/21
lim_(x→∞) x(tan^(−1) (((x+1)/(x+4)))−(π/4)) =?
$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{4}}\right)−\frac{\pi}{\mathrm{4}}\right)\:=?\: \\ $$
Answered by gsk2684 last updated on 08/Jul/21
lim_(x→∞) x(tan^(−1) (((((x+1)/(x+4))−1)/(1+((x+1)/(x+4)))))) lim_(x→∞) x(tan^(−1) (((−3)/(2x+5)))) lim_(x→∞) x(((−3)/(2x+5)))=−(3/2)
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{{x}+\mathrm{1}}{{x}+\mathrm{4}}−\mathrm{1}}{\mathrm{1}+\frac{{x}+\mathrm{1}}{{x}+\mathrm{4}}}\right)\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{3}}{\mathrm{2}{x}+\mathrm{5}}\right)\right) \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left(\frac{−\mathrm{3}}{\mathrm{2}{x}+\mathrm{5}}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
f(x)=x(arctan(((x+1)/(x+4)))−(π/4)) ⇒f(x)=_(x=(1/t)) (1/t)(arctan(((1+(1/t))/(4+(1/t))))−(π/4)) =(1/t)(arctan(((t+1)/(4t+1)))−(π/4)) =z(twe have arctan(((t+1)/(4t+1)))=arctan((1/4)(((4t+1+3)/(4t+1)))=arctan((1/4)+(3/(4(4t+1)))) ∼arctan((1/4)+(3/4)(1−4t))=arctan(1−3t) we have tan(arctan(1−3t)−(π/4))=((1−3t−1)/(1+1−3t))=((−3t)/(2−3t)) =>z(t)∼(1/t)arctan(((−3t)/((2−3t))))∼=(1/t)((−3t)/(2−3t))→−(3/2)(t→0) ⇒ lim_(x→∞) f(x)=−(3/2)
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\left(\mathrm{arctan}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{4}}\right)−\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=_{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}} \:\frac{\mathrm{1}}{\mathrm{t}}\left(\mathrm{arctan}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}}}{\mathrm{4}+\frac{\mathrm{1}}{\mathrm{t}}}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{t}}\left(\mathrm{arctan}\left(\frac{\mathrm{t}+\mathrm{1}}{\mathrm{4t}+\mathrm{1}}\right)−\frac{\pi}{\mathrm{4}}\right)\:=\mathrm{z}\left(\mathrm{twe}\:\mathrm{have}\right. \\ $$$$\mathrm{arctan}\left(\frac{\mathrm{t}+\mathrm{1}}{\mathrm{4t}+\mathrm{1}}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4t}+\mathrm{1}+\mathrm{3}}{\mathrm{4t}+\mathrm{1}}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{4t}+\mathrm{1}\right)}\right)\right. \\ $$$$\sim\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{4t}\right)\right)=\mathrm{arctan}\left(\mathrm{1}−\mathrm{3t}\right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{tan}\left(\mathrm{arctan}\left(\mathrm{1}−\mathrm{3t}\right)−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}−\mathrm{3t}−\mathrm{1}}{\mathrm{1}+\mathrm{1}−\mathrm{3t}}=\frac{−\mathrm{3t}}{\mathrm{2}−\mathrm{3t}} \\ $$$$=>\mathrm{z}\left(\mathrm{t}\right)\sim\frac{\mathrm{1}}{\mathrm{t}}\mathrm{arctan}\left(\frac{−\mathrm{3t}}{\left(\mathrm{2}−\mathrm{3t}\right)}\right)\sim=\frac{\mathrm{1}}{\mathrm{t}}\frac{−\mathrm{3t}}{\mathrm{2}−\mathrm{3t}}\rightarrow−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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