Question Number 174915 by cortano1 last updated on 14/Aug/22
$$\:{Find}\:{the}\:{value}\:{of}\:{a}\:{for}\:{which}\: \\ $$$$\:{the}\:{limit}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({ax}\right)−\mathrm{arctan}\:{x}−{x}}{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} } \\ $$$${is}\:{finite}\:{and}\:{then}\:{evaluate}\:{the}\:{limit} \\ $$
Answered by Ar Brandon last updated on 14/Aug/22
![lim_(x→0) ((sin(ax)−arctanx−x)/(x^3 +x^4 )) =lim_(x→0) (((ax−(((ax)^3 )/6))−(x−(x^3 /3))−x)/(x^3 +x^4 )) [a−1−1=0 ⇒a=2] =lim_(x→0) ((−(8/6)x^3 +(1/3)x^3 )/(x^3 +x^4 ))=lim_(x→0) ((−x^3 )/(x^3 (1+x))) =lim_(x→0) ((−1)/(1+x))=−1](Q174919.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left({ax}\right)−\mathrm{arctan}{x}−{x}}{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} } \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({ax}−\frac{\left({ax}\right)^{\mathrm{3}} }{\mathrm{6}}\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)−{x}}{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{a}−\mathrm{1}−\mathrm{1}=\mathrm{0}\:\Rightarrow{a}=\mathrm{2}\right] \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{8}}{\mathrm{6}}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} \left(\mathrm{1}+{x}\right)} \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{1}}{\mathrm{1}+{x}}=−\mathrm{1} \\ $$