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Question Number 31528 by abdo imad last updated on 09/Mar/18

find lim_(x→2) ((x^2 −2^x )/(arctanx −artan2)) .

$${find}\:{lim}_{{x}\rightarrow\mathrm{2}} \:\:\frac{{x}^{\mathrm{2}} \:\:−\mathrm{2}^{{x}} }{{arctanx}\:−{artan}\mathrm{2}}\:. \\ $$

Commented by abdo imad last updated on 12/Mar/18

let put f(x)=2^x =e^(xln2) we have f^′ (x)=ln(2).2^x f(x)=f(2) +((x−2)/(1!)) f^′ (2) + (((x−2)^2 )/(2!))ξ(x) with_(x→2) ξ(x)→0 f(x)=4 +(x−2)ln2 .2^2 +(((x−2)^2 )/(2!)) ξ(x)⇒ x^2 −f(x)=x^2 −4 −ln(2)(x−2) 2^2 −(((x−2)^2 )/(2!)) ξ(x) =(x−2)( x+2 −ln2 .2^2 −(1/(2!))ξ(x)) let put g(x)=arctanx we have g(x)= g(2) +((x−2)/(1!))g^′ (2) +(((x−2)^2 )/(2!)) θ(x) with θ(x)_(x→2) →0 we have g^′ (x)= (1/(1+x^2 )) ⇒g^′ (2)= (1/5) ⇒ g(x)=arctan2 +(1/5)((x−2)/(1!)) +(((x−2)^2 )/(2!)) θ(x)⇒ arctanx −arctan2= (x−2)( (1/5) +((x−2)/(2!)) θ(x)) ⇒ lim_(x→2) ((x^2 −2^x )/(arctanx −arctan2)) =lim_(x→2) ((x+2 −4ln2 −(1/2)ξ(x))/((1/5) +((x−2)/(2!))θ(x))) =5(4 −4 ln2) .

$${let}\:{put}\:{f}\left({x}\right)=\mathrm{2}^{{x}} \:={e}^{{xln}\mathrm{2}} \:{we}\:{have}\:{f}^{'} \left({x}\right)={ln}\left(\mathrm{2}\right).\mathrm{2}^{{x}} \\ $$$${f}\left({x}\right)={f}\left(\mathrm{2}\right)\:+\frac{{x}−\mathrm{2}}{\mathrm{1}!}\:{f}^{'} \left(\mathrm{2}\right)\:+\:\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}\xi\left({x}\right)\:{with}_{{x}\rightarrow\mathrm{2}} \:\xi\left({x}\right)\rightarrow\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{4}\:\:+\left({x}−\mathrm{2}\right){ln}\mathrm{2}\:.\mathrm{2}^{\mathrm{2}} \:+\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}\:\xi\left({x}\right)\Rightarrow \\ $$$${x}^{\mathrm{2}} \:−{f}\left({x}\right)={x}^{\mathrm{2}} \:−\mathrm{4}\:\:−{ln}\left(\mathrm{2}\right)\left({x}−\mathrm{2}\right)\:\mathrm{2}^{\mathrm{2}} \:−\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}\:\xi\left({x}\right) \\ $$$$=\left({x}−\mathrm{2}\right)\left(\:{x}+\mathrm{2}\:−{ln}\mathrm{2}\:.\mathrm{2}^{\mathrm{2}} \:\:−\frac{\mathrm{1}}{\mathrm{2}!}\xi\left({x}\right)\right)\:{let}\:{put}\:{g}\left({x}\right)={arctanx} \\ $$$${we}\:{have}\:{g}\left({x}\right)=\:{g}\left(\mathrm{2}\right)\:+\frac{{x}−\mathrm{2}}{\mathrm{1}!}{g}^{'} \left(\mathrm{2}\right)\:+\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}\:\theta\left({x}\right)\:{with} \\ $$$$\theta\left({x}\right)_{{x}\rightarrow\mathrm{2}} \rightarrow\mathrm{0}\:\:{we}\:{have}\:{g}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow{g}^{'} \left(\mathrm{2}\right)=\:\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow \\ $$$${g}\left({x}\right)={arctan}\mathrm{2}\:\:+\frac{\mathrm{1}}{\mathrm{5}}\frac{{x}−\mathrm{2}}{\mathrm{1}!}\:+\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}\:\theta\left({x}\right)\Rightarrow \\ $$$${arctanx}\:−{arctan}\mathrm{2}=\:\left({x}−\mathrm{2}\right)\left(\:\frac{\mathrm{1}}{\mathrm{5}}\:+\frac{{x}−\mathrm{2}}{\mathrm{2}!}\:\theta\left({x}\right)\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{2}} \:\:\:\frac{{x}^{\mathrm{2}} \:−\mathrm{2}^{{x}} }{{arctanx}\:−{arctan}\mathrm{2}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{2}} \:\:\:\:\:\frac{{x}+\mathrm{2}\:−\mathrm{4}{ln}\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\xi\left({x}\right)}{\frac{\mathrm{1}}{\mathrm{5}}\:+\frac{{x}−\mathrm{2}}{\mathrm{2}!}\theta\left({x}\right)}\:=\mathrm{5}\left(\mathrm{4}\:−\mathrm{4}\:{ln}\mathrm{2}\right)\:. \\ $$$$\:\:\:\:\:\: \\ $$

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