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lim-x-0-cosx-logx-




Question Number 128456 by Study last updated on 07/Jan/21
lim_(x→0) (cosx)^(logx) =???
$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =??? \\ $$
Commented by mr W last updated on 07/Jan/21
→1^(−∞) =1
$$\rightarrow\mathrm{1}^{−\infty} =\mathrm{1} \\ $$
Commented by Study last updated on 07/Jan/21
what is the practic??
$${what}\:{is}\:{the}\:{practic}?? \\ $$
Answered by bobhans last updated on 07/Jan/21
L=lim_(x→0) (cos x)^(ln x)     ln L=lim_(x→0) ln x.ln (cos x)  ln L=lim_(x→0) ((ln (cos x))/(1/(ln (x))))=lim_(x→0)  ((−tan x)/((((−1)/(x ln^2 (x))))))  ln L=lim_(x→0) x.tan x.ln^2 (x)=0  L=e^0 =1
$$\mathcal{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:{x}\right)^{\mathrm{ln}\:{x}} \:\: \\ $$$$\mathrm{ln}\:\mathcal{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}ln}\:{x}.\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right) \\ $$$$\mathrm{ln}\:\mathcal{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)}{\frac{\mathrm{1}}{\mathrm{ln}\:\left({x}\right)}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:{x}}{\left(\frac{−\mathrm{1}}{{x}\:\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)}\right)} \\ $$$$\mathrm{ln}\:\mathcal{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}.\mathrm{tan}\:{x}.\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)=\mathrm{0} \\ $$$$\mathcal{L}={e}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by mr W last updated on 08/Jan/21
but lim_(x→0) x.tan x.ln^2 (x)=0∙0∙∞ =^(?)  0
$${but}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}.\mathrm{tan}\:{x}.\mathrm{ln}\:^{\mathrm{2}} \left({x}\right)=\mathrm{0}\centerdot\mathrm{0}\centerdot\infty\:\overset{?} {=}\:\mathrm{0} \\ $$

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