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Question Number 194250 by horsebrand11 last updated on 01/Jul/23
find the value of a for which the limit lim_(x→0) ((sin (ax)−tan^(−1) (x)−x)/(x^3 +x^4 )) is finite and then evaluate the limit
$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{ax}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} }\:\mathrm{is}\:\mathrm{finite}\: \\ $$$$\:\mathrm{and}\:\mathrm{then}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{limit}\: \\ $$
Answered by qaz last updated on 01/Jul/23
sin (ax)=ax−(1/6)a^3 x^3 +... ,arctan x=x−(1/3)x^3 +... lim_(x→0) (((a−2)x+((1/3)−(1/6)a^3 )x^3 )/x^3 )=−1 ,a=2
$$\mathrm{sin}\:\left({ax}\right)={ax}−\frac{\mathrm{1}}{\mathrm{6}}{a}^{\mathrm{3}} {x}^{\mathrm{3}} +…\:\:\:\:\:,\mathrm{arctan}\:{x}={x}−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +… \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({a}−\mathrm{2}\right){x}+\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}{a}^{\mathrm{3}} \right){x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }=−\mathrm{1}\:\:\:,{a}=\mathrm{2} \\ $$
Answered by gatocomcirrose last updated on 02/Jul/23
lim_(x→0) ((acos(ax)−(1/(1+x^2 ))−1)/(3x^2 +4x^3 ))=((a−2)/0)→±∞ a=2⇒lim_(x→0) ((−a^2 sen(ax)+((2x)/((1+x^2 )^2 )))/(6x+12x^2 )) =lim_(x→0) ((−a^3 cos(ax)+((2(1+x^2 )^2 −8x^2 (1+x^2 ))/((1+x^2 )^4 )))/(6+24x)) =((−a^3 +2)/6)=((−8+2)/6)=−1
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{acos}\left(\mathrm{ax}\right)−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} }=\frac{\mathrm{a}−\mathrm{2}}{\mathrm{0}}\rightarrow\pm\infty \\ $$$$\mathrm{a}=\mathrm{2}\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{a}^{\mathrm{2}} \mathrm{sen}\left(\mathrm{ax}\right)+\frac{\mathrm{2x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{6x}+\mathrm{12x}^{\mathrm{2}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{a}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{ax}\right)+\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} }}{\mathrm{6}+\mathrm{24x}} \\ $$$$=\frac{−\mathrm{a}^{\mathrm{3}} +\mathrm{2}}{\mathrm{6}}=\frac{−\mathrm{8}+\mathrm{2}}{\mathrm{6}}=−\mathrm{1} \\ $$

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