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Question Number 87243 by john santu last updated on 03/Apr/20
lim_(x→0) ((x− sin x)/( (√((1−cos x)^p )))) = k k = constant , find p
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\:\mathrm{sin}\:{x}}{\:\sqrt{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{{p}} }}\:=\:{k}\: \\ $$$${k}\:=\:{constant}\:,\:\mathrm{find}\:\mathrm{p}\: \\ $$
Commented by jagoll last updated on 04/Apr/20
lim_(x→0) ((x−(x−(x^3 /6)+o(x^3 )))/( (√((2sin^2 ((x/2)))^p )))) = k (1/(6×2^(p/2) )) × lim_(x→0) ((x^3 −o(x^3 ))/(sin^((2p)/2) ((x/2)))) = k since k = constant , so ((2p)/2) should be p = 3 . such that k = (8/(6×2(√2))) = (2/(3(√2))) = ((2(√2))/(3×2)) = ((√2)/3)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)\right)}{\:\sqrt{\left(\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)^{\mathrm{p}} }}\:=\:\mathrm{k} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}×\mathrm{2}^{\frac{\mathrm{p}}{\mathrm{2}}} }\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{sin}\:^{\frac{\mathrm{2p}}{\mathrm{2}}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:=\:\mathrm{k} \\ $$$$\mathrm{since}\:\mathrm{k}\:=\:\mathrm{constant}\:,\:\mathrm{so}\:\frac{\mathrm{2p}}{\mathrm{2}} \\ $$$$\mathrm{should}\:\mathrm{be}\:\mathrm{p}\:=\:\mathrm{3}\:.\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{8}}{\mathrm{6}×\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}×\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Answered by mr W last updated on 04/Apr/20
=lim_(x→0) ((1−cos x)/((p/2)(1−cos x)^((p/2)−1) (sin x))) =(2/p)lim_(x→0) (((1−cos x)^(2−(p/2)) )/((sin x))) =(2/p)lim_(x→0) (((2sin^2 (x/2))^(2−(p/2)) )/(2 cos (x/2)sin (x/2))) =±(2^(2−(p/2)) /p)lim_(x→0) (((sin (x/2))^(3−p) )/(cos (x/2)))= { ((=0 or ∞, if 3−p≠0)),((±(2^(2−(p/2)) /p)=k, if p=3)) :} ⇒p=3, k=±(2^(2−(p/2)) /p)=±((√2)/3)
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\frac{{p}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\frac{{p}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{sin}\:{x}\right)} \\ $$$$=\frac{\mathrm{2}}{{p}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)^{\mathrm{2}−\frac{{p}}{\mathrm{2}}} }{\left(\mathrm{sin}\:{x}\right)} \\ $$$$=\frac{\mathrm{2}}{{p}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}−\frac{{p}}{\mathrm{2}}} }{\mathrm{2}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mathrm{sin}\:\frac{{x}}{\mathrm{2}}} \\ $$$$=\pm\frac{\mathrm{2}^{\mathrm{2}−\frac{{p}}{\mathrm{2}}} }{{p}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)^{\mathrm{3}−{p}} }{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}=\begin{cases}{=\mathrm{0}\:{or}\:\infty,\:{if}\:\mathrm{3}−{p}\neq\mathrm{0}}\\{\pm\frac{\mathrm{2}^{\mathrm{2}−\frac{{p}}{\mathrm{2}}} }{{p}}={k},\:{if}\:{p}=\mathrm{3}}\end{cases} \\ $$$$\Rightarrow{p}=\mathrm{3},\:{k}=\pm\frac{\mathrm{2}^{\mathrm{2}−\frac{{p}}{\mathrm{2}}} }{{p}}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Commented by jagoll last updated on 04/Apr/20
why (d/dx) [(√((1−cos x)^p )) ] = (p/2)(1−cos x)^((p/2)−1) (−sin x) ? not (p/2) (1−cos x)^((p/2)−1) (+ sin x) ? mr w?
$$\mathrm{why}\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{p}} }\:\right]\:= \\ $$$$\frac{\mathrm{p}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\frac{\mathrm{p}}{\mathrm{2}}−\mathrm{1}} \:\left(−\mathrm{sin}\:\mathrm{x}\right)\:? \\ $$$$\mathrm{not}\:\frac{\mathrm{p}}{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\frac{\mathrm{p}}{\mathrm{2}}−\mathrm{1}} \:\left(+\:\mathrm{sin}\:\mathrm{x}\right)\:?\: \\ $$$$\mathrm{mr}\:\mathrm{w}? \\ $$
Commented by mr W last updated on 04/Apr/20
+sin x is correct. but lim_(x→0^+ ) ∗∗=k=+((√2)/3) lim_(x→0^− ) ∗∗=k=−((√2)/3)
$$+\mathrm{sin}\:{x}\:{is}\:{correct}. \\ $$$${but} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\ast\ast={k}=+\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\ast\ast={k}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$
Commented by john santu last updated on 04/Apr/20
ok sir
$$\mathrm{ok}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 04/Apr/20
since k=0 is also constant, the answer should be p≤3.
$${since}\:{k}=\mathrm{0}\:{is}\:{also}\:{constant},\:{the}\: \\ $$$${answer}\:{should}\:{be}\:{p}\leqslant\mathrm{3}. \\ $$

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