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Question Number 228358 by Math1 last updated on 09/Apr/26
Find: lim_(x→(𝛑/2)) [(x − (π/2)) tanx] = ?
$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\mathrm{lim}}\:\left[\left(\mathrm{x}\:−\:\frac{\pi}{\mathrm{2}}\right)\:\mathrm{tanx}\right]\:=\:? \\ $$
Answered by cherokeesay last updated on 09/Apr/26
Answered by Lara2440 last updated on 10/Apr/26
lim_(x→(π/2)) (x−(π/2))∙tan(x)=lim_(x→(π/2)) (((x−(π/2))sin(x))/(cos(x))) Let t →^(substitute) x−(π/2) lim_(t→0) ((t∙sin(t+(π/2)))/(cos(t+(π/2))))=−lim_(t→0) ((t∙cos(t))/(sin(t))) sin(t)<t<tan(t) , t∈(0,(π/2)) .....(A)^(∗∗) sin(t)<t<((sin(t))/(cos(t))) ⇒ sin(t)cos(t)<t∙cos(t)<((sin(t))/(cos(t)))∙cos(t) =sin(t)cos(t)∙(1/(sin(t)))<((t∙cos(t))/(sin(t)))<((sin(t))/(sin(t))) =cos(t)<((t∙cos(t))/(sin(t)))<1 ∴ lim_(t→0) cos(t)<lim_(t→0) ((t∙cos(t))/(sin(t)))<lim_(t→0) 1 1<lim_(t→0) ((t∙cos(t))/(sin(t)))<1 ⇒ ∴−lim_(t→0) ((t∙cos(t))/(sin(t)))=−1 But can we really sure lim_(t→0) cos(t)=1....?? and lim_(x→a) A<lim_(x→a) B<lim_(x→a) C and iff ⇒ lim_(x→a) A=L ,lim_(x→a) C=L lim_(x→a) B=L...?? Ans) proof... lim_(t→0) cos(t)=1 for all 𝛆>0 there exist 𝛅>0 , whenever t∈R\{0} such that 0<∣t∣<𝛅 ⇒^(Implies) ∣cos(x)−1∣<𝛆 sin^2 ((t/2))=((1−cos(t))/2) (from the half angle formula) 1−cos(t)=2∙sin^2 ((t/2)) sin(t)<t .....(A)^(∗∗) t∈(0,(π/2)) ∴∣(1−cos(t))∣=2∙sin^2 ((t/2))<(t^2 /2) 𝛅=min((π/2),(√(2𝛆))) choose 𝛆=0.00001 , 𝛅=min((π/2),(√(2∙(0.00001)))) 𝛅≈0.004472136.... ∣1−cos(𝛅)∣<𝛆 ...??? ∣(1−cos(𝛅))∣≈0.00000968<0.00001 True!! ∴ lim_(t→0) cos(t)=1 Suppose that the inequality A<B<C holds for any x in a given ingerval x∈E assuming that the limit operation preserves this Order and that lim_(x→a) A , lim_(x→a) B , lim_(x→a) C all converge. Q) would lim_(x→a) B=L hold if lim_(x→a) A=L and lim_(x→a) C=L...?? for a totally ordered set (X,≤) the following three binary relations are Satisfied 1. transitive relation ∀a,b,c∈X ,aRb∧bRc ⇒ aRc ex). a≤b , b≤c ⇒ a≤c 2. antisymmetric relation ∀a,b∈X , aRb∧bRa ⇒ a=b ex). a≤b , b≤a ⇒ a=b 3. total relation ∀x,y∈X , (x,y)∈R or (y,x)∈R for a each function A,B,C inequality A<B<C holds, and the limit operation preserves this Order and funtion A,B,C converge to a single value. and Let lim_(x→a) A=L<lim_(x→a) B and lim_(x→a) B<lim_(x→a) C=L ∴ lim_(x→a) B=L (by antisymmetric relation)
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)\centerdot\mathrm{tan}\left({x}\right)=\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\left({x}−\frac{\pi}{\mathrm{2}}\right)\mathrm{sin}\left({x}\right)}{\mathrm{cos}\left({x}\right)} \\ $$$$\mathrm{Let}\:{t}\:\overset{\mathrm{substitute}} {\rightarrow}{x}−\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}\centerdot\mathrm{sin}\left({t}+\frac{\pi}{\mathrm{2}}\right)}{\mathrm{cos}\left({t}+\frac{\pi}{\mathrm{2}}\right)}=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)} \\ $$$$\: \\ $$$$\mathrm{sin}\left({t}\right)<{t}<\mathrm{tan}\left({t}\right)\:,\:{t}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:…..\left({A}\right)^{\ast\ast} \\ $$$$\mathrm{sin}\left({t}\right)<{t}<\frac{\mathrm{sin}\left({t}\right)}{\mathrm{cos}\left({t}\right)}\:\Rightarrow\:\mathrm{sin}\left({t}\right)\mathrm{cos}\left({t}\right)<{t}\centerdot\mathrm{cos}\left({t}\right)<\frac{\mathrm{sin}\left({t}\right)}{\mathrm{cos}\left({t}\right)}\centerdot\mathrm{cos}\left({t}\right) \\ $$$$=\mathrm{sin}\left({t}\right)\mathrm{cos}\left({t}\right)\centerdot\frac{\mathrm{1}}{\mathrm{sin}\left({t}\right)}<\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}<\frac{\mathrm{sin}\left({t}\right)}{\mathrm{sin}\left({t}\right)} \\ $$$$=\mathrm{cos}\left({t}\right)<\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}<\mathrm{1} \\ $$$$\therefore\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\left({t}\right)<\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}<\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{1} \\ $$$$\mathrm{1}<\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}<\mathrm{1}\:\Rightarrow\:\therefore−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}\centerdot\mathrm{cos}\left({t}\right)}{\mathrm{sin}\left({t}\right)}=−\mathrm{1} \\ $$$$\: \\ $$$$\mathrm{But}\:\mathrm{can}\:\mathrm{we}\:\mathrm{really}\:\mathrm{sure}\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\left({t}\right)=\mathrm{1}….?? \\ $$$$\mathrm{and} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{A}<\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}<\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{C}\:\mathrm{and}\:\mathrm{iff}\:\Rightarrow\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{A}={L}\:,\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{C}={L} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}={L}…?? \\ $$$$\: \\ $$$$\left.{Ans}\right) \\ $$$$\mathrm{proof}… \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\left({t}\right)=\mathrm{1}\: \\ $$$$\mathrm{for}\:\mathrm{all}\:\boldsymbol{\epsilon}>\mathrm{0}\:\mathrm{there}\:\mathrm{exist}\:\boldsymbol{\delta}>\mathrm{0}\:,\:\mathrm{whenever}\:{t}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{0}<\mid{t}\mid<\boldsymbol{\delta}\:\overset{\mathrm{Implies}} {\Rightarrow}\:\mid\mathrm{cos}\left({x}\right)−\mathrm{1}\mid<\boldsymbol{\epsilon} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)=\frac{\mathrm{1}−\mathrm{cos}\left({t}\right)}{\mathrm{2}}\:\left(\mathrm{from}\:\mathrm{the}\:\mathrm{half}\:\mathrm{angle}\:\mathrm{formula}\right) \\ $$$$\mathrm{1}−\mathrm{cos}\left({t}\right)=\mathrm{2}\centerdot\mathrm{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\left({t}\right)<{t}\:…..\left({A}\right)^{\ast\ast} \:\:{t}\in\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$$$\therefore\mid\left(\mathrm{1}−\mathrm{cos}\left({t}\right)\right)\mid=\mathrm{2}\centerdot\mathrm{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)<\frac{{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\boldsymbol{\delta}=\mathrm{min}\left(\frac{\pi}{\mathrm{2}},\sqrt{\mathrm{2}\boldsymbol{\epsilon}}\right) \\ $$$$\mathrm{choose}\:\boldsymbol{\epsilon}=\mathrm{0}.\mathrm{00001}\:,\:\boldsymbol{\delta}=\mathrm{min}\left(\frac{\pi}{\mathrm{2}},\sqrt{\mathrm{2}\centerdot\left(\mathrm{0}.\mathrm{00001}\right)}\right) \\ $$$$\boldsymbol{\delta}\approx\mathrm{0}.\mathrm{004472136}…. \\ $$$$\mid\mathrm{1}−\mathrm{cos}\left(\boldsymbol{\delta}\right)\mid<\boldsymbol{\epsilon}\:…??? \\ $$$$\mid\left(\mathrm{1}−\mathrm{cos}\left(\boldsymbol{\delta}\right)\right)\mid\approx\mathrm{0}.\mathrm{00000968}<\mathrm{0}.\mathrm{00001}\:\:\mathrm{True}!! \\ $$$$\: \\ $$$$\therefore\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}\left({t}\right)=\mathrm{1} \\ $$$$\:\: \\ $$$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{inequality}\:{A}<{B}<{C}\:\mathrm{holds} \\ $$$$\mathrm{for}\:\mathrm{any}\:{x}\:\mathrm{in}\:\mathrm{a}\:\mathrm{given}\:\mathrm{ingerval}\:{x}\in\mathbb{E} \\ $$$$\mathrm{assuming}\:\mathrm{that}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{operation}\:\mathrm{preserves}\:\mathrm{this}\:\mathrm{Order} \\ $$$$\mathrm{and}\:\mathrm{that}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{A}\:,\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}\:,\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{C}\:\mathrm{all}\:\mathrm{converge}. \\ $$$$\left.\mathcal{Q}\right) \\ $$$$\mathrm{would}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}={L}\:\mathrm{hold}\:\mathrm{if}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{A}={L}\:\mathrm{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{C}={L}…?? \\ $$$$\: \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{totally}\:\mathrm{ordered}\:\mathrm{set}\:\left({X},\leq\right)\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{three}\:\mathrm{binary}\:\mathrm{relations}\:\mathrm{are}\:\mathrm{Satisfied} \\ $$$$\: \\ $$$$\mathrm{1}.\:\mathrm{transitive}\:\mathrm{relation} \\ $$$$\forall{a},{b},{c}\in{X}\:,{a}\mathcal{R}{b}\wedge{b}\mathcal{R}{c}\:\Rightarrow\:{a}\mathcal{R}{c} \\ $$$$\left.\mathrm{ex}\right).\:{a}\leq{b}\:,\:{b}\leq{c}\:\Rightarrow\:{a}\leq{c} \\ $$$$\: \\ $$$$\mathrm{2}.\:\mathrm{antisymmetric}\:\mathrm{relation} \\ $$$$\forall{a},{b}\in{X}\:,\:{a}\mathcal{R}{b}\wedge{b}\mathcal{R}{a}\:\Rightarrow\:{a}={b} \\ $$$$\left.\mathrm{ex}\right).\:{a}\leq{b}\:,\:{b}\leq{a}\:\Rightarrow\:{a}={b} \\ $$$$\: \\ $$$$\mathrm{3}.\:\mathrm{total}\:\mathrm{relation} \\ $$$$\forall{x},{y}\in{X}\:,\:\left({x},{y}\right)\in\mathcal{R}\:\mathrm{or}\:\left({y},{x}\right)\in\mathcal{R} \\ $$$$\: \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{each}\:\mathrm{function}\:{A},{B},{C}\:\mathrm{inequality}\:{A}<{B}<{C}\:\mathrm{holds}, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{operation}\:\mathrm{preserves}\:\mathrm{this}\:\mathrm{Order} \\ $$$$\: \\ $$$$\mathrm{and}\:\mathrm{funtion}\:{A},{B},{C}\:\mathrm{converge}\:\mathrm{to}\:\mathrm{a}\:\mathrm{single}\:\mathrm{value}. \\ $$$$\: \\ $$$$\mathrm{and}\:\mathrm{Let} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{A}={L}<\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}\:\:\mathrm{and}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}<\underset{{x}\rightarrow{a}} {\mathrm{lim}}{C}={L}\: \\ $$$$\therefore\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{B}={L}\: \\ $$$$\left(\mathrm{by}\:\mathrm{antisymmetric}\:\mathrm{relation}\right) \\ $$
Answered by Ghisom_ last updated on 10/Apr/26
lim_(x→π/2) ((x−π/2)tan x) = =−lim_(t→0) (t/(tan t)) = [l′Ho^� pital] =−lim_(t→0) (1/(1/cos^2 t)) =−lim_(t→0) cos^2 t =−1
$$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\left(\left({x}−\pi/\mathrm{2}\right)\mathrm{tan}\:{x}\right)\:= \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{t}}{\mathrm{tan}\:{t}}\:= \\ $$$$\:\:\:\:\:\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right] \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}/\mathrm{cos}^{\mathrm{2}} \:{t}}\:=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{cos}^{\mathrm{2}} \:{t}\:=−\mathrm{1} \\ $$

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