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lim-x-0-cosx-logx-




Question Number 129387 by Study last updated on 15/Jan/21
lim_(x→0) (cosx)^(logx) =???
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =??? \\ $$
Commented by Study last updated on 15/Jan/21
who will solve?
$${who}\:{will}\:{solve}? \\ $$
Answered by mindispower last updated on 15/Jan/21
=e^(ln(x)ln(cos(x)))   ln(cos(x))=ln(1−(x^2 /2)+o(x^2 ))=−(x^2 /2)+o(x^2 )  =e^(ln(x)(−(x^2 /2)+o(x^2 )))   lim_(x→0) x^2 ln(x)=0  we get e^0 =1
$$={e}^{{ln}\left({x}\right){ln}\left({cos}\left({x}\right)\right)} \\ $$$${ln}\left({cos}\left({x}\right)\right)={ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$={e}^{{ln}\left({x}\right)\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}^{\mathrm{2}} {ln}\left({x}\right)=\mathrm{0} \\ $$$${we}\:{get}\:{e}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by Study last updated on 16/Jan/21
i think to undefined?
$${i}\:{think}\:{to}\:{undefined}? \\ $$

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