Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 114543 by bemath last updated on 19/Sep/20

 lim_(x→0)  ((x^2  arctan (2x))/(x cos x+tan x)) ?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{arctan}\:\left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:? \\ $$

Answered by bobhans last updated on 19/Sep/20

lim_(x→0)  ((x^2  tan^(−1) (2x))/(x cos x+tan x)) = lim_(x→0)   ((2x^3 )/(x(1−(x^2 /2))+(x+(1/3)x^3 )))  = lim_(x→0)  ((2x^3 )/(2x−(1/6)x^3 )) = lim_(x→0)  ((2x^2 )/(2−(1/6)x^2 )) = 0

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{2}{x}^{\mathrm{3}} }{{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)+\left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 19/Sep/20

by hospital theorem  we get  L =lim_(x→0)        ((2x arctanx+x^2 .(2/(1+4x^2 )))/(cosx −xsinx +1+tan^2 x)) =(0/2)=0

$$\mathrm{by}\:\mathrm{hospital}\:\mathrm{theorem}\:\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{L}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\:\:\:\frac{\mathrm{2x}\:\mathrm{arctanx}+\mathrm{x}^{\mathrm{2}} .\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }}{\mathrm{cosx}\:−\mathrm{xsinx}\:+\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}\:=\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com