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 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)

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)+\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)+\mathrm{2sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}}\:\right].\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}}{\mathrm{sin}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}+\mathrm{cos}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{px}}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{p}}\:×\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{p}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com