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Question Number 115780 by bemath last updated on 28/Sep/20
lim_(x→0) (cos x)^(1/x^2 ) = ? lim_(x→∞) (e^(3x) −5x)^(1/x) =? lim_(x→0) ((e^(2x) −2e^x +1)/(cos 3x−2cos 2x+cos x))=? lim_(x→0) (1/x^2 )−cot^2 x =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({e}^{\mathrm{3}{x}} −\mathrm{5}{x}\right)^{\frac{\mathrm{1}}{{x}}} \:=? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} +\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{2cos}\:\mathrm{2}{x}+\mathrm{cos}\:{x}}=? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{cot}\:^{\mathrm{2}} {x}\:=? \\ $$
Commented by bobhans last updated on 28/Sep/20
L=lim_(x→0) (cos x)^(1/x^2 ) = e^(lim_(x→0) (cos x−1).(1/x^2 )) L=e^(lim_(x→0) ((1−(1/2)x^2 −1)/x^2 )) = e^(−(1/2)) = (1/( (√e)))
$${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}−\mathrm{1}\right).\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$${L}={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Sep/20
lim_(x→0) (cosx)^(1/x^2 ) =(1−(x^2 /2))^(−(2/x^2 ).−(1/2)) =e^(−(1/2)) = (1/( (√e)))
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} =\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }.−\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\ $$
Answered by bobhans last updated on 28/Sep/20
L=lim_(x→∞) (e^(3x) −5x)^(1/x) ln L=lim_(x→∞) ((ln (e^(3x) −5x))/x) ln L=lim_(x→∞) ((3e^(3x) −5)/(e^(3x) −5x)) = lim_(x→∞) ((9e^(3x) )/(3e^(3x) −5)) ln L=lim_(x→∞) ((9e^(3x) )/(e^(3x) (3−(5/e^(3x) )))) = 3 L= e^3
$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({e}^{\mathrm{3}{x}} −\mathrm{5}{x}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:\left({e}^{\mathrm{3}{x}} −\mathrm{5}{x}\right)}{{x}} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}{e}^{\mathrm{3}{x}} −\mathrm{5}}{{e}^{\mathrm{3}{x}} −\mathrm{5}{x}}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{9}{e}^{\mathrm{3}{x}} }{\mathrm{3}{e}^{\mathrm{3}{x}} −\mathrm{5}} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{9}{e}^{\mathrm{3}{x}} }{{e}^{\mathrm{3}{x}} \left(\mathrm{3}−\frac{\mathrm{5}}{{e}^{\mathrm{3}{x}} }\right)}\:=\:\mathrm{3} \\ $$$${L}=\:{e}^{\mathrm{3}} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Sep/20
lim_(x→0) (1/x^2 )−((cos^2 x)/(sin^2 x))=((sin^2 x−x^2 cos^2 x)/(x^2 sin^2 x))=((x^2 (1−(x^2 /6))^2 −x^2 (1−(x^2 /2))^2 )/(x^2 sin^2 x)) =(((2−(x^2 /2)−(x^2 /6))((x^2 /2)−(x^2 /6)))/x^2 )=((x^2 (2−((2x^2 )/3))((1/3)))/x^2 )=(2/3)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{\left(\mathrm{2}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\right)}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{2}−\frac{\mathrm{2x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by Dwaipayan Shikari last updated on 28/Sep/20
lim_(x→0) (((e^x −1)^2 )/x^2 ).(x^2 /((cos3x+cosx−2cos2x))) =1.1.(x^2 /(2(cos2xcosx−cos2x)))=(x^2 /(2cos2x(cosx−1))) =−(1/2).(x^2 /(2sin^2 (x/2)))=−(1/4).(x^2 /(x^2 /4))=−1
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{e}^{\mathrm{x}} −\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }.\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{cos3x}+\mathrm{cosx}−\mathrm{2cos2x}\right)} \\ $$$$=\mathrm{1}.\mathrm{1}.\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{cos2xcosx}−\mathrm{cos2x}\right)}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2cos2x}\left(\mathrm{cosx}−\mathrm{1}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}}=−\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{x}^{\mathrm{2}} }{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}}=−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
(cosx)^(1/x^2 ) =e^((1/x^2 )ln(cosx)) we have ln(cosx)∼ln(1−(x^2 /2))∼−(x^2 /2) ⇒ (1/x^2 )ln(cosx)∼−(1/2) ⇒lim_(x→0) (cosx)^(1/x^2 ) =e^(−(1/2)) =(1/( (√e)))
$$\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)} \:\:\mathrm{we}\:\mathrm{have}\:\mathrm{ln}\left(\mathrm{cosx}\right)\sim\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\sim−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{cosx}\right)\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }} \:=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
we have (e^(3x) −5x)^(1/x) =(e^(3x) (1−5x e^(−3x) )^(1/x) =e^3 (1−5xe^(−3x) )^(1/x) =e^3 e^((1/x)ln(1−5xe^(−3x) )) ∼e^3 e^((1/x)(−5xe^(−3x) )) =e^3 e^(−5e^(−3x) ) ⇒e^(−2) =(1/e^2 ) ⇒ lim_(x→0) (e^(3x) −5x)^(1/x) =(1/e^2 )
$$\mathrm{we}\:\mathrm{have}\:\left(\mathrm{e}^{\mathrm{3x}} −\mathrm{5x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:=\left(\mathrm{e}^{\mathrm{3x}} \left(\mathrm{1}−\mathrm{5x}\:\mathrm{e}^{−\mathrm{3x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:=\mathrm{e}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{5xe}^{−\mathrm{3x}} \right)^{\frac{\mathrm{1}}{\mathrm{x}}} \right. \\ $$$$=\mathrm{e}^{\mathrm{3}} \:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}−\mathrm{5xe}^{−\mathrm{3x}} \right)} \:\sim\mathrm{e}^{\mathrm{3}} \:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}\left(−\mathrm{5xe}^{−\mathrm{3x}} \right)} \:=\mathrm{e}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{5e}^{−\mathrm{3x}} } \:\Rightarrow\mathrm{e}^{−\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\left(\mathrm{e}^{\mathrm{3x}} −\mathrm{5x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:=\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} } \\ $$
Commented by Dwaipayan Shikari last updated on 28/Sep/20
e^3 e^(−5e^(−3x) ) =e^3 .e^(−5(0)) =e^3 (As e^(−3x) →0)
$$\mathrm{e}^{\mathrm{3}} \mathrm{e}^{−\mathrm{5e}^{−\mathrm{3x}} } =\mathrm{e}^{\mathrm{3}} .\mathrm{e}^{−\mathrm{5}\left(\mathrm{0}\right)} =\mathrm{e}^{\mathrm{3}} \:\:\:\:\:\:\left(\mathrm{As}\:\:\mathrm{e}^{−\mathrm{3x}} \rightarrow\mathrm{0}\right) \\ $$
Commented by Bird last updated on 28/Sep/20
soory i take limit at 0 lim_(x→+∞) (e^(3x) −5x)^(1/x) =e^3 .e^0 =e^3
$${soory}\:{i}\:{take}\:{limit}\:{at}\:\mathrm{0} \\ $$$${lim}_{{x}\rightarrow+\infty} \left({e}^{\mathrm{3}{x}} −\mathrm{5}{x}\right)^{\frac{\mathrm{1}}{{x}}} \:={e}^{\mathrm{3}} .{e}^{\mathrm{0}} \:={e}^{\mathrm{3}} \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
let f(x) =((e^(2x) −2e^x +1)/(cos(3x)−2cos(2x)+cosx)) (x→0) ⇒ f(x) ∼((1+2x+2x^2 −2(1+x+(x^2 /2))+1)/(1−((9x^2 )/2)−2(1−2x^2 )+1−(x^2 /2))) ⇒f(x)∼ (x^2 /(−((9x^2 )/2)+4x^2 −(x^2 /2))) =(x^2 /(−5x^(2 ) +4x^2 )) =(x^2 /(−x^2 )) =−1 ⇒ lim_(x→0) f(x) =−1
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} \:+\mathrm{1}}{\mathrm{cos}\left(\mathrm{3x}\right)−\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cosx}}\:\:\left(\mathrm{x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:\sim\frac{\mathrm{1}+\mathrm{2x}+\mathrm{2x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\mathrm{x}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)+\mathrm{1}}{\mathrm{1}−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)+\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\:\frac{\mathrm{x}^{\mathrm{2}} }{−\frac{\mathrm{9x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4x}^{\mathrm{2}} −\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}}\:=\frac{\mathrm{x}^{\mathrm{2}} }{−\mathrm{5x}^{\mathrm{2}\:} +\mathrm{4x}^{\mathrm{2}} }\:=\frac{\mathrm{x}^{\mathrm{2}} }{−\mathrm{x}^{\mathrm{2}} }\:=−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)\:=−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
let use hospital we have lim_(x→0) ((e^(2x) −2e^x +1)/(cos(3x)−2cos(2x)+cosx)) =lim_(x→0) ((2e^(2x) −2e^x )/(−3sin(3x)+4sin(2x)−sinx)) =lim_(x→0) ((4e^(2x) −2e^x )/(−9cos(3x)+8cos(2x)−cosx)) =(2/(−9+8−1)) =(2/(−2))=−1
$$\mathrm{let}\:\mathrm{use}\:\mathrm{hospital}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} \:+\mathrm{1}}{\mathrm{cos}\left(\mathrm{3x}\right)−\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cosx}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{2e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} }{−\mathrm{3sin}\left(\mathrm{3x}\right)+\mathrm{4sin}\left(\mathrm{2x}\right)−\mathrm{sinx}} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{4e}^{\mathrm{2x}} −\mathrm{2e}^{\mathrm{x}} }{−\mathrm{9cos}\left(\mathrm{3x}\right)+\mathrm{8cos}\left(\mathrm{2x}\right)−\mathrm{cosx}}\:=\frac{\mathrm{2}}{−\mathrm{9}+\mathrm{8}−\mathrm{1}}\:=\frac{\mathrm{2}}{−\mathrm{2}}=−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 28/Sep/20
we have (1/x^2 )−cotan^2 x =(1/x^2 )−((cos^2 x)/(sin^2 x)) =((sin^2 x−x^2 cos^2 x)/(x^2 sin^2 x)) =((((1−cos(2x))/2)−x^2 ×((1+cos(2x))/2))/(x^2 .((1−cos(2x))/2))) =(1/2).((1−cos(2x)−x^2 (1+cos(2x)))/(x^2 (1−cos(2x)))) ⇒(1/x^2 )−cotan^2 x ∼(1/2) ((2x^2 −x^2 (1+1−2x^2 ))/(x^2 ×(2x^2 )))=(1/2)×((2x^4 )/(2x^4 )) ⇒ lim_(x→0) (1/x^2 )−cotan^2 x =(1/2)
$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{cotan}^{\mathrm{2}} \mathrm{x}\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\frac{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}−\mathrm{x}^{\mathrm{2}} ×\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} .\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)−\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)\right)} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{cotan}^{\mathrm{2}} \mathrm{x}\:\sim\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{1}−\mathrm{2x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} ×\left(\mathrm{2x}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2x}^{\mathrm{4}} }{\mathrm{2x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{cotan}^{\mathrm{2}} \mathrm{x}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 28/Sep/20
Answered by john santu last updated on 29/Sep/20
lim_(x→0) (1/x^2 )−(1/(tan^2 x)) = lim_(x→0) ((tan^2 x−x^2 )/(x^2 tan^2 x)) = lim_(x→0) ((tan x+x)/(tan x)) ×lim_(x→0) ((tan x−x)/(x^2 tan x)) = lim_(x→0) (1+(x/(tan x))) ×lim_(x→0) ((x+(1/3)x^3 −x)/x^3 ) = 2 × lim_(x→0) (((1/3)x^3 )/x^3 ) = (2/3)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:^{\mathrm{2}} {x}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:{x}+{x}}{\mathrm{tan}\:{x}}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{2}} \:\mathrm{tan}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{{x}}{\mathrm{tan}\:{x}}\right)\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} −{x}}{{x}^{\mathrm{3}} } \\ $$$$=\:\mathrm{2}\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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