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Question Number 161248 by cortano last updated on 15/Dec/21

  lim_(x→0)  (((1+sin^3 x)^4 −(1+tan^3 x)^4 )/x^5 ) =?

$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{3}} {x}\right)^{\mathrm{4}} −\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{3}} {x}\right)^{\mathrm{4}} }{{x}^{\mathrm{5}} }\:=?\: \\ $$

Answered by bobhans last updated on 15/Dec/21

 L = 2×2×lim_(x→0)  ((tan^3 (x)(cos^3 x−1))/x^5 )   L = 12×lim_(x→0)  ((cos x−1)/x^2 )   L = 12×lim_(x→0)  ((−2sin^2 ((x/2)))/x^2 ) = −6

$$\:\mathcal{L}\:=\:\mathrm{2}×\mathrm{2}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{3}} \left(\mathrm{x}\right)\left(\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{1}\right)}{\mathrm{x}^{\mathrm{5}} } \\ $$$$\:\mathcal{L}\:=\:\mathrm{12}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\:\mathcal{L}\:=\:\mathrm{12}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}^{\mathrm{2}} }\:=\:−\mathrm{6}\: \\ $$

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