Menu Close

lim-x-0-1-sin-x-cos-x-1-sin-px-cos-px-




Question Number 134171 by liberty last updated on 28/Feb/21
 lim_(x→0)  ((1+sin x−cos x)/(1+sin (px)−cos (px))) =?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\left(\mathrm{px}\right)−\mathrm{cos}\:\left(\mathrm{px}\right)}\:=? \\ $$
Answered by malwan last updated on 28/Feb/21
lim_(x→0)  ((1+(x−...)−(1−...))/(1+(px−...)−(1−...)))  = (1/p)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mathrm{1}+\left({x}−…\right)−\left(\mathrm{1}−…\right)}{\mathrm{1}+\left({px}−…\right)−\left(\mathrm{1}−…\right)}\:\:=\:\frac{\mathrm{1}}{{p}} \\ $$
Answered by physicstutes last updated on 28/Feb/21
 L = lim_(x→0) (((cos x + sin x)/(p cos px + p sin px))) = (1/p)
$$\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cos}\:{x}\:+\:\mathrm{sin}\:{x}}{{p}\:\mathrm{cos}\:{px}\:+\:{p}\:\mathrm{sin}\:{px}}\right)\:=\:\frac{\mathrm{1}}{{p}} \\ $$
Answered by liberty last updated on 01/Mar/21
 lim_(x→0)  ((2sin^2 ((1/2)x)+2sin ((1/2)x)cos ((1/2)x))/(2sin^2 ((1/2)px)+2sin ((1/2)px)cos ((1/2)px)))  =lim_(x→0)  [((sin (1/2)x)/(sin (1/2)px)) ]. lim_(x→0)  [((sin (1/2)x+cos (1/2)x)/(sin  (1/2)px+cos  (1/2)px)) ]  = (1/p) ×1 =(1/p)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)+\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)+\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}}\:\right].\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{sin}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}+\mathrm{cos}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{p}}\:×\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{p}} \\ $$

Leave a Reply

Your email address will not be published.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