Menu Close

lim-x-0-x-3-1-x-sin-3-x-1-2-x-3-tan-x-1-2x-3-1-3-1-ln-1-x-2-

Tinku Tara 0 Comments
Pin




Question Number 168144 by cortano1 last updated on 04/Apr/22
lim_(x→0) ((x^3 (√(1+x)) −sin^3 x −(1/2)x^3 tan x)/((((1+2x^3 ))^(1/3) −1)ln (1+x^2 )))=?
$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \sqrt{\mathrm{1}+{x}}\:−\mathrm{sin}\:^{\mathrm{3}} {x}\:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \:\mathrm{tan}\:{x}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{2}{x}^{\mathrm{3}} }−\mathrm{1}\right)\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}=?\:\:\:\:\:\:\:\: \\ $$
Answered by qaz last updated on 05/Apr/22
lim_(x→0) ((x^3 (√(1+x))−sin^3 x−(1/2)x^3 tan x)/((((1+2x^3 ))^(1/3) −1)ln(1+x^2 ))) =lim_(x→0) ((x^3 (1+(1/2)x−(1/8)x^2 +...)−x^3 (1−(1/6)x^2 ...)^3 −(1/2)x^3 (x+...))/((2/3)x^5 )) =lim_(x→0) (((1+(1/2)x−(1/8)x^2 +...)−(1+ ((3),(1) )(−(1/6)x^2 )+...)−(1/2)(x+...))/((2/3)x^2 )) =lim_(x→0) (((−(1/8)+(1/2))x^2 +o(x^2 ))/((2/3)x^2 )) =(9/(16))
$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{1}+\mathrm{x}}−\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} \mathrm{tan}\:\mathrm{x}}{\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{2x}^{\mathrm{3}} }−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} +…\right)−\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{2}} …\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}+…\right)}{\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{5}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} +…\right)−\left(\mathrm{1}+\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\left(−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{2}} \right)+…\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}+…\right)}{\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{9}}{\mathrm{16}} \\ $$
Commented by cortano1 last updated on 05/Apr/22
ans = (9/(16))
$${ans}\:=\:\frac{\mathrm{9}}{\mathrm{16}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *