Menu Close

find-lim-x-0-ln-2-cos-2x-ln-1-xsin-3x-




Question Number 73330 by mathmax by abdo last updated on 10/Nov/19
find lim_(x→0)    ((ln(2−cos(2x)))/(ln(1+xsin(3x))))
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right)}{{ln}\left(\mathrm{1}+{xsin}\left(\mathrm{3}{x}\right)\right)} \\ $$
Commented by mathmax by abdo last updated on 11/Nov/19
let f(x)=((ln(2−cos(2x)))/(ln(1+xsin(3x)))) we have cos(2x)∼1−(((2x)^2 )/2)=1−2x^2  ⇒  2−cos(2x)∼2−1+2x^2 =1+2x^2   and ln(1+2x^2 )∼2x^2   (x ∼0) ⇒  also xsin(3x) ∼x(3x)=3x^2  ⇒f(x)∼((2x^2 )/(3x^2 ))  ⇒lim_(x→0) f(x)=(2/3)
$${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right)}{{ln}\left(\mathrm{1}+{xsin}\left(\mathrm{3}{x}\right)\right)}\:{we}\:{have}\:{cos}\left(\mathrm{2}{x}\right)\sim\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\sim\mathrm{2}−\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:\:{and}\:{ln}\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)\sim\mathrm{2}{x}^{\mathrm{2}} \:\:\left({x}\:\sim\mathrm{0}\right)\:\Rightarrow \\ $$$${also}\:{xsin}\left(\mathrm{3}{x}\right)\:\sim{x}\left(\mathrm{3}{x}\right)=\mathrm{3}{x}^{\mathrm{2}} \:\Rightarrow{f}\left({x}\right)\sim\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}{x}^{\mathrm{2}} }\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by Smail last updated on 10/Nov/19
ln(2−cos(2x))=ln(1+2sin^2 (x))∼_0 2sin^2 (x)  ln(1+xsin(3x))∼_0 xsin(3x)  lim_(x→0) ((ln(2−cos(2x)))/(ln(1+xsin(3x))))=lim_(x→0) ((2sin^2 x)/(xsin(3x)))  =lim_(x→0) 2((sinx)/x)×((sinx)/x)×((3x)/(sin(3x)))×(1/3)  =(2/3)
$${ln}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right)={ln}\left(\mathrm{1}+\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right)\right)\underset{\mathrm{0}} {\sim}\mathrm{2}{sin}^{\mathrm{2}} \left({x}\right) \\ $$$${ln}\left(\mathrm{1}+{xsin}\left(\mathrm{3}{x}\right)\right)\underset{\mathrm{0}} {\sim}{xsin}\left(\mathrm{3}{x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right)}{{ln}\left(\mathrm{1}+{xsin}\left(\mathrm{3}{x}\right)\right)}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{2}{sin}^{\mathrm{2}} {x}}{{xsin}\left(\mathrm{3}{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\mathrm{2}\frac{{sinx}}{{x}}×\frac{{sinx}}{{x}}×\frac{\mathrm{3}{x}}{{sin}\left(\mathrm{3}{x}\right)}×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Leave a Reply

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