Question Number 175880 by cortano1 last updated on 08/Sep/22
Answered by blackmamba last updated on 09/Sep/22
![lim_(x→0) ((bx^2 (((sin x)/x))^2 +3x(((tan 3x)/(3x)))+2x^2 (((sin (1/2)x)/x))^2 )/( (√2) x(((sin (3/2)x)/x))+x^(2b) (((sin 3x)/x))^(2b) )) = lim_(x→0) [ ((bx^2 +3x+(1/2)x^2 )/(((3(√2))/2)x +(9x)^(2b) )) ] = lim_(x→0) [((bx+3+(1/2)x)/(((3(√2))/2)+9^(2b) .x^(2b−1) )) ] = ((2.3)/(3(√2))) = (√2)](https://www.tinkutara.com/question/Q175897.png)
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{bx}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:{x}}{{x}}\right)^{\mathrm{2}} +\mathrm{3}{x}\left(\frac{\mathrm{tan}\:\mathrm{3}{x}}{\mathrm{3}{x}}\right)+\mathrm{2}{x}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{{x}}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}\:{x}\left(\frac{\mathrm{sin}\:\frac{\mathrm{3}}{\mathrm{2}}{x}}{{x}}\right)+{x}^{\mathrm{2}{b}} \left(\frac{\mathrm{sin}\:\mathrm{3}{x}}{{x}}\right)^{\mathrm{2}{b}} }\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{{bx}^{\mathrm{2}} +\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} }{\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}{x}\:+\left(\mathrm{9}{x}\right)^{\mathrm{2}{b}} }\:\right] \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{{bx}+\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}{x}}{\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{9}^{\mathrm{2}{b}} .{x}^{\mathrm{2}{b}−\mathrm{1}} }\:\right] \\ $$$$\:\:=\:\frac{\mathrm{2}.\mathrm{3}}{\mathrm{3}\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{2}}\: \\ $$