lim-x-0-1-x-1-sin-x-

Question Number 88092 by ar247 last updated on 08/Apr/20

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

Commented by ar247 last updated on 08/Apr/20

$${please}\:{help} \\ $$

Commented by jagoll last updated on 08/Apr/20

Commented by jagoll last updated on 08/Apr/20

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

Answered by Rio Michael last updated on 08/Apr/20

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

Commented by ar247 last updated on 08/Apr/20

$${how} \\ $$

Commented by Rio Michael last updated on 08/Apr/20

this is an informal idea lim_(x→0^+ ) ((1/x)−(1/(sin x))) = lim_(x→0^+ ) (1/x) − lim_(x→0^+ ) (1/(sin x)) = +∞ − ∞ = −∞ (1/x) < (1/(sin x )) as x → 0^+

$$\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{informal}\:\mathrm{idea}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\:−\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:=\:+\infty\:−\:\infty\:=\:−\infty \\ $$$$\frac{\mathrm{1}}{{x}}\:<\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}\:}\:\:\mathrm{as}\:{x}\:\rightarrow\:\mathrm{0}^{+} \\ $$

Commented by Joel578 last updated on 08/Apr/20

Be careful! ∞−∞ is an undetermined form, so we can′t conclude that ∞−∞ = −∞ lim_(x→0^+ ) ((1/x) − (1/(sin x))) = lim_(x→0^+ ) ((sin x − x)/(x sin x)) = lim_(x→0^+ ) ((cos x − 1)/(sin x + x cos x)) = lim_(x→0^+ ) ((−sin x)/(2cos x − x sin x)) = 0

$$\mathrm{Be}\:\mathrm{careful}!\:\:\:\:\infty−\infty\:\mathrm{is}\:\mathrm{an}\:\mathrm{undetermined}\:\mathrm{form}, \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{conclude}\:\mathrm{that}\:\infty−\infty\:=\:−\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}\:−\:{x}}{{x}\:\mathrm{sin}\:{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}\:−\:\mathrm{1}}{\mathrm{sin}\:{x}\:+\:{x}\:\mathrm{cos}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{−\mathrm{sin}\:{x}}{\mathrm{2cos}\:{x}\:−\:{x}\:\mathrm{sin}\:{x}}\:=\:\mathrm{0} \\ $$

Commented by Ar Brandon last updated on 08/Apr/20

$${Angle}\:{is}\:{in}\:{radians}\:{Mr}\:{Michael}. \\ $$

Answered by Ar Brandon last updated on 08/Apr/20

⇒lim_(x→0^+ ) (((sin x−x)/(xsin x)))=lim_(x→0^+ ) (((d/dx)(sin x−x))/((d(xsin x))/dx)) ⇒lim_(x→0^+ ) (((cos x−1)/(xcos x+sin x)))=(0/0) ⇒lim_(x→0^+ ) (((−sin x)/(−xsin x+cos x+cos x)))=−(0/2)=0

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

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