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Question Number 68178 by Mikael last updated on 06/Sep/19

$$\underset{x\to 0}{lim}\frac{{\left(cosx\right)}^{sin2x}-1}{x^3}=?$$

Commented by mathmax by abdo last updated on 06/Sep/19

$$letf\left(x\right)=\frac{{\left(cosx\right)}^{sin\left(2x\right)}-1}{x^3}wehavef\left(x\right)=\frac{e^{sin\left(2x\right)ln\left(cosx\right)}-1}{x^3}$$$$wehavecosx\sim 1-\frac{x^2}{2}\Rightarrow ln\left(cosx\right)\sim ln(1-\frac{x^2}{2})$$$${ln}^{{}^{\prime }}(1-u)=\frac{-1}{1-u}=-(1+u+o\left(u\right))\Rightarrow$$$$ln(1-u)=-u-\frac{u^2}{2}+o\left(u^2\right)\Rightarrow ln(1-\frac{x^2}{2})=-\frac{x^2}{2}-\frac{x^4}{4}+o\left(x^4\right)$$$$alsosin\left(2x\right)\sim 2x\Rightarrow sin\left(2x\right)ln\left(cosx\right)\sim 2x(-\frac{x^2}{2}-\frac{x^4}{4})$$$$=-x^3-\frac{1}{2}{x}^5\Rightarrow e^{sin\left(2x\right)ln\left(cosx\right)}\sim e^{-x^3-\frac{1}{2}{x}^5}\sim 1-x^3-\frac{1}{2}{x}^5\Rightarrow$$$$f\left(x\right)\sim \frac{-x^3-\frac{1}{2}{x}^5}{x^3}\Rightarrow f\left(x\right)\sim -1-\frac{1}{2}{x}^2\Rightarrow {lim}_{x\to 0}f\left(x\right)=-1$$

Commented by Mikael last updated on 06/Sep/19

$$thankyou.$$

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