Question Number 20138 by tammi last updated on 22/Aug/17
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{ax}}{\mathrm{1}−\mathrm{cos}\:{bx}} \\ $$$$ \\ $$
Answered by ajfour last updated on 22/Aug/17
![=lim_(x→0) ((2sin^2 (((ax)/2)))/(2sin^2 (((bx)/2))))=(a^2 /b^2 )lim_(x→0) [((sin (((ax)/2)))/((((ax)/2))))×(((((bx)/2)))/(sin (((bx)/2))))]^2 =(a^2 /b^2 )×lim_(x→0) [((sin (ax/2))/(ax/2))]÷lim_(x→0) [((sin (bx/2))/(bx/2))] =(a^2 /b^2 )×1÷1 =(a^2 /b^2 ) .](Q20144.png)
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{ax}}{\mathrm{2}}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{bx}}{\mathrm{2}}\right)}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{sin}\:\left(\frac{{ax}}{\mathrm{2}}\right)}{\left(\frac{{ax}}{\mathrm{2}}\right)}×\frac{\left(\frac{{bx}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{bx}}{\mathrm{2}}\right)}\right]^{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{sin}\:\left({ax}/\mathrm{2}\right)}{{ax}/\mathrm{2}}\right]\boldsymbol{\div}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{sin}\:\left({bx}/\mathrm{2}\right)}{{bx}/\mathrm{2}}\right] \\ $$$$=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }×\mathrm{1}\boldsymbol{\div}\mathrm{1}\:=\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }\:. \\ $$
Commented by tammi last updated on 23/Aug/17
$${thanks}\:{a}\:{lottt} \\ $$