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Question Number 169004 by mathlove last updated on 23/Apr/22

lim_(x→(π/2)) (((sinx−1)/(x−(π/2))))=?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{{sinx}−\mathrm{1}}{{x}−\frac{\pi}{\mathrm{2}}}\right)=? \\ $$

Commented by infinityaction last updated on 23/Apr/22

(let)p = lim_(x→(π/2)) (((sin x−sin (π/2))/(x−(π/2)))) p = lim_(x→(π/2)) {((2cos (((x+π/2)/2))sin (((x−π/2)/2)))/(2(((x−π/2)/2))))} p = cos (π/2) p = 0

$$\:\:\:\:\:\:\:\:\:\:\:\left({let}\right){p}\:\:\:\:\:\:\:\:=\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:{x}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}}{{x}−\frac{\pi}{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\:\:\:\:=\:\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left\{\frac{\mathrm{2cos}\:\left(\frac{{x}+\pi/\mathrm{2}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}−\pi/\mathrm{2}}{\mathrm{2}}\right)}{\mathrm{2}\left(\frac{{x}−\pi/\mathrm{2}}{\mathrm{2}}\right)}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}\:\:\:\:=\:\:\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{p}\:\:\:\:\:\:=\:\:\:\:\mathrm{0} \\ $$$$ \\ $$

Commented by mathlove last updated on 23/Apr/22

sir sinp+sinq=2cos((p+q)/2)∙sin((p−q)/2)

$${sir} \\ $$$${sinp}+{sinq}=\mathrm{2}{cos}\frac{{p}+{q}}{\mathrm{2}}\centerdot{sin}\frac{{p}−{q}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 23/Apr/22

lim_(x→(π/2)) (((sinx−1)/(x−(π/2)))) =lim_(x→(π/2)) (((sinx−sin (π/2))/(x−(π/2)))) =(sin x)′∣_(x=(π/2)) =(cos x)∣_(x=(π/2)) =0

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

Answered by qaz last updated on 23/Apr/22

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

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

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