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Question Number 106810 by Study last updated on 07/Aug/20

lim_(x→0) ((tanx−sinx)/(sinx(cos2x−cosx)))=???

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{tanx}−{sinx}}{{sinx}\left({cos}\mathrm{2}{x}−{cosx}\right)}=??? \\ $$

Answered by bemath last updated on 07/Aug/20

@bemath@ lim_(x→0) ((tan x−sin x)/(sin x(cos 2x−cos x))) ? lim_(x→0) ((tan x(1−cos x))/(sin x(−2sin (((3x)/2))sin ((x/2))))) = lim_(x→0) ((tan x(2sin^2 ((x/2)))/(−2sin x sin (((3x)/2))sin ((x/2))))= −1×lim_(x→0) ((tan x.sin ((x/2)))/(sin x sin (((3x)/2))))=−(1/3)

$$\:\:\:\:\:@\mathrm{bemath}@ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}\left(\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{x}\right)}\:? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{sin}\:\mathrm{x}\left(−\mathrm{2sin}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right.}{−\mathrm{2sin}\:\mathrm{x}\:\mathrm{sin}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}= \\ $$$$−\mathrm{1}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}.\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

lim_(x→0) ((sinx((1/(cosx))−1))/(sinx(−2sin3xsinx))) lim_(x→0) (((2sin^2 (x/2))/(cosx))/(−2sin((3x)/2)sin(x/2)))=lim_(x→0) (((2((x/2))^2 )/(cosx))/(−2.3x^2 )).4=−(1/3) sinx→x

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinx}\left(\frac{\mathrm{1}}{{cosx}}−\mathrm{1}\right)}{{sinx}\left(−\mathrm{2}{sin}\mathrm{3}{xsinx}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{cosx}}}{−\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }{{cosx}}}{−\mathrm{2}.\mathrm{3}{x}^{\mathrm{2}} }.\mathrm{4}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${sinx}\rightarrow{x} \\ $$

Answered by 1549442205PVT last updated on 07/Aug/20

lim_(x→0) ((tanx−sinx)/(sinx(cos2x−cosx)))=lim_(x→0) ((sinx((1/(cosx))−1))/(sinx(cos2x−cosx))) =lim_(x→0) (((1/(cosx))−1)/(cos2x−cosx)))=lim_(x→0) ((1−cosx)/(2cos^2 x−cosx−1)) =lim _(x→0) ((1−cosx)/((cosx−1)(2cosx+1)))=lim _(x→0) ((−1)/(2cosx+1)) =−(1/3)

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{tanx}−{sinx}}{{sinx}\left({cos}\mathrm{2}{x}−{cosx}\right)}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sinx}\left(\frac{\mathrm{1}}{\mathrm{cosx}}−\mathrm{1}\right)}{\mathrm{sinx}\left(\mathrm{cos2x}−\mathrm{cosx}\right)} \\ $$$$={li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{cosx}}−\mathrm{1}}{\left.{cos}\mathrm{2}{x}−{cosx}\right)}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}−\mathrm{cosx}−\mathrm{1}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}\:}\frac{\mathrm{1}−\mathrm{cosx}}{\left(\mathrm{cosx}−\mathrm{1}\right)\left(\mathrm{2cosx}+\mathrm{1}\right)}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}\:}\frac{−\mathrm{1}}{\mathrm{2cosx}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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