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Question Number 167617 by mathlove last updated on 21/Mar/22

(1)lim_(x→0) [tan((π/4)−x)]^(cotx) =? (2)lim_(x→0) [(1/(sin x))−(1/x)]=?

$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]^{{cotx}} =? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{1}}{{x}}\right]=? \\ $$

Answered by cortano1 last updated on 21/Mar/22

(2) lim_(x→0) (((x−sin x)/(xsin x))) = lim_(x→0) (((x−sin x)/(x^2 (((sin x)/x))))) = lim_(x→0) (((x−sin x)/x^2 ))=lim_(x→0) (((1−cos x)/(2x))) = 0

$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{xsin}\:\mathrm{x}}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)}\right) \\ $$$$\:\:\:\:\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{2x}}\right) \\ $$$$\:\:\:\:=\:\mathrm{0}\: \\ $$

Answered by LEKOUMA last updated on 21/Mar/22

(2) lim_(x→0) ((1/(sin x))−(1/x)) lim_(x→0) ((x/(xsin x))−((sin x)/(xsin x)))=lim_(x→0) (((x−sin x)/(xsin x))) Hospital lim_(x→0) ((((x−sin x)′)/((xsin x)′)))=lim_(x→0) ((((x)′−(sin x)′)/((xsin x)′))) lim_(x→0 ) ((1−cos x)/(sin x+xcos x))=lim_(x→0 ) (((1−cos x)/x)/(((sin x)/x)+cos x))=(0/(1+1))=(0/2)=0 lim_(x→0) ((1/(sin x))−(1/x))=0

$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}}{{x}\mathrm{sin}\:{x}}−\frac{\mathrm{sin}\:{x}}{{x}\mathrm{sin}\:{x}}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}−\mathrm{sin}\:{x}}{{x}\mathrm{sin}\:{x}}\right) \\ $$$${Hospital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\left({x}−\mathrm{sin}\:{x}\right)'}{\left({x}\mathrm{sin}\:{x}\right)'}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\left({x}\right)'−\left(\mathrm{sin}\:{x}\right)'}{\left({x}\mathrm{sin}\:{x}\right)'}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}\:\:} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}+{x}\mathrm{cos}\:{x}}=\underset{{x}\rightarrow\mathrm{0}\:\:} {\mathrm{lim}}\frac{\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}}}{\frac{\mathrm{sin}\:{x}}{{x}}+\mathrm{cos}\:{x}}=\frac{\mathrm{0}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{0} \\ $$

Answered by qaz last updated on 21/Mar/22

lim_(x→0) [tan ((π/4)−x)]^(cot x) =lim_(x→0) (((cot x−1)/(cot x+1)))^(cot x) =lim_(x→0) [(1−(2/(cot x+1)))^((cot x+1)/2) ]^2 (1−(2/(cot x+1)))^(−1) =e^(−2)

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)\right]^{\mathrm{cot}\:\mathrm{x}} =\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{cot}\:\mathrm{x}−\mathrm{1}}{\mathrm{cot}\:\mathrm{x}+\mathrm{1}}\right)^{\mathrm{cot}\:\mathrm{x}} =\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{cot}\:\mathrm{x}+\mathrm{1}}\right)^{\frac{\mathrm{cot}\:\mathrm{x}+\mathrm{1}}{\mathrm{2}}} \right]^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{cot}\:\mathrm{x}+\mathrm{1}}\right)^{−\mathrm{1}} =\mathrm{e}^{−\mathrm{2}} \\ $$

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