Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)

$$\mathrm{L}=\underset{\mathrm{x}\rightarrow\pi} {\mathrm{lim}}\frac{\mathrm{cos}\left(\frac{\pi−\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{cos}\left(\mathrm{sin}\left(\pi−\mathrm{x}\right)\right)−\mathrm{1}} \\ $$$$\mathrm{let}\:\mathrm{t}=\pi−\mathrm{x},\:\mathrm{if}\:\mathrm{x}\rightarrow\pi,\:\Rightarrow\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\frac{\mathrm{t}}{\mathrm{2}}−\mathrm{1}}{\mathrm{cos}\left(\mathrm{sint}\right)−\mathrm{1}} \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}−\mathrm{cos}\frac{\mathrm{t}}{\mathrm{2}}}{\left(\frac{\mathrm{t}}{\mathrm{2}}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{sint}\right)}{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}×\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{t}^{\mathrm{2}} }} \\ $$$$\mathrm{L}=\frac{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com