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Question Number 159646 by cortano last updated on 19/Nov/21

     lim_(x→0)  ((1−(cos x)^(sin x) )/x^3 ) =?

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

Answered by FongXD last updated on 19/Nov/21

L=lim_(x→0) ((1−(cosx)^(sinx) )/x^3 )     form: (0/0)  L=−lim_(x→0) ((e^(sinxln(cosx)) −1)/(sinxln(cosx)))×((sinx)/x)×((ln(1+cosx−1))/(cosx−1))×((1−cosx)/x^2 )(−1)  L=−1×1×1×(1/2)(−1)=(1/2)

$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\left(\mathrm{cosx}\right)^{\mathrm{sinx}} }{\mathrm{x}^{\mathrm{3}} }\:\:\:\:\:\mathrm{form}:\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\mathrm{L}=−\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{sinxln}\left(\mathrm{cosx}\right)} −\mathrm{1}}{\mathrm{sinxln}\left(\mathrm{cosx}\right)}×\frac{\mathrm{sinx}}{\mathrm{x}}×\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{cosx}−\mathrm{1}\right)}{\mathrm{cosx}−\mathrm{1}}×\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }\left(−\mathrm{1}\right) \\ $$$$\mathrm{L}=−\mathrm{1}×\mathrm{1}×\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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