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A-lim-x-0-1-cos2x-2x-2-lim-x-0-2sin-2-x-2x-2-lim-x-0-sinx-x-2-1-B-lim-x-0-1-xcotx-lim-x-0-tanx-x-lim-x-0-sinx-x-1-cosx-1-




Question Number 205138 by Thokna last updated on 10/Mar/24
A=lim_(x→0  ) ((1−cos2x)/(2x^2 ))      =lim_(x→0) ((2sin^2 x)/(2x^2 ))      =lim_(x→0) (((sinx)/x))^2 =1  B=lim_(x→0)  (1/(xcotx))      =lim_(x→0)  ((tanx)/x)=lim_(x→0) ((sinx)/x)×(1/(cosx))=1
$${A}=\underset{{x}\rightarrow\mathrm{0}\:\:} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sin}^{\mathrm{2}} {x}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}{x}}{{x}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${B}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}\mathrm{cot}{x}} \\ $$$$\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}{x}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}{x}}{{x}}×\frac{\mathrm{1}}{\mathrm{cos}{x}}=\mathrm{1} \\ $$

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