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Question Number 129510 by Adel last updated on 16/Jan/21

$$\underset{x\to 0}{\lim }{\left(\cos \mathrm{x}\right)}^{\log \mathrm{x}}=?$$

Commented by mr W last updated on 16/Jan/21

$$\Rightarrow Q129387$$

Answered by greg_ed last updated on 16/Jan/21

$${\left(\mathbf{cos}\mathit{x}\right)}^{\mathit{l}\mathit{n}\mathit{x}}=\mathbf{cos}{}^{\mathit{l}\mathit{n}\mathit{x}}\mathit{x}=\mathit{e}{}^{\mathit{l}\mathit{n}\mathit{x}\mathit{l}\mathit{n}\left(\mathbf{cos}\mathit{x}\right)}$$$${\mathbf{let}}^{\prime }\mathbf{s}\mathbf{rewrite}\mathbf{this}\mathit{u}\left(\mathit{x}\right)=\mathit{l}\mathit{n}\mathit{x}\mathit{l}\mathit{n}\left(\mathbf{cos}\mathit{x}\right).$$$$\mathit{u}\left(\mathit{x}\right)=\frac{\mathit{l}\mathit{n}\mathit{x}}{\frac{1}{\mathit{l}\mathit{n}\left(\mathbf{cos}\mathit{x}\right)}}$$$$$$$$1.\mathbf{Applying}{\mathbf{L}}^{\prime }{\mathbf{Hopital}}^{\prime }\mathbf{s}\mathbf{rule},\mathbf{we}\mathbf{get}$$$$\mathit{u}\left(\mathit{x}\right)=\frac{\mathit{l}\mathit{n}{}^2\left(\mathbf{cos}\mathit{x}\right)}{\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$$$$2.\mathbf{Applying}{\mathbf{L}}^{\prime }{\mathbf{Hopital}}^{\prime }\mathbf{s}\mathbf{rule},\mathbf{we}\mathbf{get}$$$$\mathit{u}\left(\mathit{x}\right)=-\frac{2\mathit{l}\mathit{n}\left(\mathbf{cos}\mathit{x}\right)\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{\mathit{t}\mathit{a}\mathit{n}\mathit{x}+\mathit{x}\mathit{s}\mathit{e}\mathit{c}{}^2\mathit{x}}$$$$3.\mathbf{Applying}{\mathbf{L}}^{\prime }{\mathbf{Hopital}}^{\prime }\mathbf{s}\mathbf{rule},\mathbf{we}\mathbf{get}$$$$\mathit{u}\left(\mathit{x}\right)=-\frac{-\mathit{t}\mathit{a}\mathit{n}{}^2\mathit{x}+\mathbf{sec}{}^2\mathit{x}\mathit{l}\mathit{n}\left(\mathbf{cos}\mathit{x}\right)}{\mathbf{sec}{}^2\mathit{x}(1+\mathit{x}\mathit{t}\mathit{a}\mathit{n}\mathit{x})}$$$$\mathbf{Plugging}\mathit{x}=0\mathbf{in}\mathbf{this},\mathbf{we}\mathbf{get}$$$$\underset{x\to 0}{\lim }\mathit{u}\left(\mathit{x}\right)=0$$$$\mathbf{So},\mathbf{finally}$$$$\underset{\mathit{u}\to 0}{\lim }\mathit{e}{}^{\mathit{u}}=1.$$$$$$

Answered by mathmax by abdo last updated on 16/Jan/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\mathrm{cosx}\right)}^{\mathrm{logx}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{logxlog}\left(\mathrm{cosx}\right)}\mathrm{we}\mathrm{have}$$$$\cos \left(\mathrm{x}\right)\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow \log \left(\mathrm{cosx}\right)\sim \log (1-\frac{{\mathrm{x}}^2}{2})\sim -\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$\mathrm{logxlog}\left(\mathrm{cosx}\right)\sim -\frac{{\mathrm{x}}^2}{2}\mathrm{logx}\to 0(\mathrm{x}\to 0)\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^0=1$$

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