Question-155639

Question Number 155639 by cortano last updated on 03/Oct/21

Commented by yeti123 last updated on 03/Oct/21

lim_(θ→0) ((sin 2θ)/(1−cos θ)) = lim_(θ→0) ((sin 2θ)/(2sin^2 (θ/2))) = lim_(θ→0) (((sin 2θ)/(2θ))×(((θ/2)^2 )/(2sin^2 (θ/2)))×((2θ)/((θ/2)^2 ))) = 1×(1/2)×lim_(θ→0) (8/θ) = ∞

$$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{cos}\:\theta}\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}\theta}×\frac{\left(\theta/\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)}×\frac{\mathrm{2}\theta}{\left(\theta/\mathrm{2}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}}{\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\infty \\ $$

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

lim_(x→0^+ ) (1/x) = ∞ lim_(x→0^− ) (1/x)=−∞

$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\infty \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{x}}=−\infty \\ $$

Commented by yeti123 last updated on 03/Oct/21

= 1×(1/2)×lim_(θ→0) (8/θ) = lim_(θ→0) (4/θ) = two-sided limit doesnt exist

$$=\:\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}}{\theta} \\ $$$$=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}}{\theta} \\ $$$$=\:\mathrm{two}-\mathrm{sided}\:\mathrm{limit}\:\mathrm{doesnt}\:\mathrm{exist} \\ $$

Answered by Ar Brandon last updated on 03/Oct/21

=lim_(ϑ→0) ((4ϑ(1−(ϑ/2)))/ϑ^2 )=lim_(x→0) ((4/ϑ)−2)→infinity

$$=\underset{\vartheta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\vartheta\left(\mathrm{1}−\frac{\vartheta}{\mathrm{2}}\right)}{\vartheta^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{4}}{\vartheta}−\mathrm{2}\right)\rightarrow\mathrm{infinity} \\ $$

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