Question Number 155639 by cortano last updated on 03/Oct/21
Commented by yeti123 last updated on 03/Oct/21
$$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{cos}\:\theta}\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}\theta}×\frac{\left(\theta/\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)}×\frac{\mathrm{2}\theta}{\left(\theta/\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}}{\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\infty \\ $$
Commented by cortano last updated on 03/Oct/21
$$\mathrm{the}\:\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}\: \\ $$
Commented by cortano last updated on 03/Oct/21
$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\infty \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}=−\infty \\ $$
Commented by yeti123 last updated on 03/Oct/21
$$=\:\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}}{\theta} \\ $$$$=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}}{\theta} \\ $$$$=\:\mathrm{two}-\mathrm{sided}\:\mathrm{limit}\:\mathrm{doesnt}\:\mathrm{exist} \\ $$
Answered by Ar Brandon last updated on 03/Oct/21
$$=\underset{\vartheta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\vartheta\left(\mathrm{1}−\frac{\vartheta}{\mathrm{2}}\right)}{\vartheta^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{4}}{\vartheta}−\mathrm{2}\right)\rightarrow\mathrm{infinity} \\ $$