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Question Number 160178 by cortano last updated on 25/Nov/21
lim_(x→π) ((sin (x/2)−1)/(cos (sin x)−1)) =?
$$\:\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)−\mathrm{1}}\:=? \\ $$
Commented by blackmamba last updated on 25/Nov/21
lim_(x→π) ((sin (x/2)−1)/(cos (sin x)−1)) = lim_(x→π) ((sin (x/2)−1)/(−2sin^2 (((sin x)/2)))) let h=sin (x/2) ;h→1 =lim_(h→1) ((h−1)/(−2sin^2 (h(√(1−h^2 ))))) = −(1/2) lim_(h→1) ((h−1)/(h^2 (1−h)(1+h))) = −(1/2)×−(1/2) = (1/4) .
$$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)−\mathrm{1}}\: \\ $$$$\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}−\mathrm{1}}{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{sin}\:{x}}{\mathrm{2}}\right)} \\ $$$$\:{let}\:{h}=\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:;{h}\rightarrow\mathrm{1}\: \\ $$$$\:\:\:\:\:\:=\underset{{h}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{h}−\mathrm{1}}{−\mathrm{2sin}\:^{\mathrm{2}} \left({h}\sqrt{\mathrm{1}−{h}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{h}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{h}−\mathrm{1}}{{h}^{\mathrm{2}} \left(\mathrm{1}−{h}\right)\left(\mathrm{1}+{h}\right)} \\ $$$$\:\:\:\:=\:−\frac{\mathrm{1}}{\mathrm{2}}×−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$
Answered by FongXD last updated on 25/Nov/21
L=lim_(x→π) ((cos(((π−x)/2))−1)/(cos(sin(π−x))−1)) let t=π−x, if x→π, ⇒ t→0 L=lim_(t→0) ((cos(t/2)−1)/(cos(sint)−1)) L=lim_(t→0) ((((1−cos(t/2))/(((t/2))^2 ))×(1/4))/(((1−cos(sint))/(sin^2 t))×((sin^2 t)/t^2 ))) L=(((1/2)×(1/4))/((1/2)×1^2 ))=(1/4)
$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{cos}\left(\frac{\pi−\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{cos}\left(\mathrm{sin}\left(\pi−\mathrm{x}\right)\right)−\mathrm{1}} \\ $$$$\mathrm{let}\:\mathrm{t}=\pi−\mathrm{x},\:\mathrm{if}\:\mathrm{x}\rightarrow\pi,\:\Rightarrow\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\frac{\mathrm{t}}{\mathrm{2}}−\mathrm{1}}{\mathrm{cos}\left(\mathrm{sint}\right)−\mathrm{1}} \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}−\mathrm{cos}\frac{\mathrm{t}}{\mathrm{2}}}{\left(\frac{\mathrm{t}}{\mathrm{2}}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{sint}\right)}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}×\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{t}^{\mathrm{2}} }} \\ $$$$\mathrm{L}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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