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Question Number 117416 by Lordose last updated on 11/Oct/20
lim_(x→∞) ((e^x^2 −cosx)/(sin^2 x))
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } −\mathrm{cosx}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$
Answered by Olaf last updated on 11/Oct/20
∀x∈R, sinx ≤ x ⇒ (1/(sin^2 x)) ≥ (1/x^2 ) e^x^2 −cosx ≥ e^x^2 −1 ((e^x^2 −cosx)/(sin^2 x)) ≥ ((e^x^2 −1)/x^2 ) But lim_(x→∞) ((e^x^2 −1)/x^2 ) = lim_(u→∞) ((e^u −1)/u) = lim_(u→∞) ((e^u −0)/1) = +∞ (Hospital′s rule) ⇒ lim_(x→∞) ((e^x^2 −cosx)/(sin^2 x)) = +∞ (by comparison)
$$\forall{x}\in\mathbb{R},\:\mathrm{sin}{x}\:\leqslant\:{x}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}}\:\geqslant\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\: \\ $$$${e}^{{x}^{\mathrm{2}} } −\mathrm{cos}{x}\:\geqslant\:{e}^{{x}^{\mathrm{2}} } −\mathrm{1} \\ $$$$\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{cos}{x}}{\mathrm{sin}^{\mathrm{2}} {x}}\:\geqslant\:\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{But}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{u}} −\mathrm{1}}{{u}}\:=\:\underset{{u}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{u}} −\mathrm{0}}{\mathrm{1}}\:=\:+\infty \\ $$$$\left(\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}\right) \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{e}^{{x}^{\mathrm{2}} } −\mathrm{cos}{x}}{\mathrm{sin}^{\mathrm{2}} {x}}\:=\:+\infty \\ $$$$\left(\mathrm{by}\:\mathrm{comparison}\right) \\ $$
Commented by Lordose last updated on 11/Oct/20
the text said (3/2)
$$\mathrm{the}\:\mathrm{text}\:\mathrm{said}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by MJS_new last updated on 11/Oct/20
obviously the text is wrong
$$\mathrm{obviously}\:\mathrm{the}\:\mathrm{text}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by Olaf last updated on 11/Oct/20
do x = nπ we have ((e^(π^2 n^2 ) −(−1)^n )/0^+ ) →∞
$${do}\:{x}\:=\:{n}\pi \\ $$$${we}\:{have}\:\frac{{e}^{\pi^{\mathrm{2}} {n}^{\mathrm{2}} } −\left(−\mathrm{1}\right)^{{n}} }{\mathrm{0}^{+} }\:\rightarrow\infty \\ $$
Commented by Olaf last updated on 11/Oct/20
(3/2) it′s in 0
$$\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{it}'\mathrm{s}\:\mathrm{in}\:\mathrm{0} \\ $$
Answered by MJS_new last updated on 11/Oct/20
−1≤cos x ≤1 0≤sin^2 x≤1 ⇒ lim =∞
$$−\mathrm{1}\leqslant\mathrm{cos}\:{x}\:\leqslant\mathrm{1} \\ $$$$\mathrm{0}\leqslant\mathrm{sin}^{\mathrm{2}} \:{x}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{lim}\:=\infty \\ $$
Answered by Bird last updated on 12/Oct/20
let f(x)=((e^x^2 −cosx)/(sin^2 x)) ⇒ f(x) =2((e^x^2 −cosx)/(1−cos(2x))) we have cosu =Σ_(n=0) ^∞ (((−1)^n u^(2n) )/((2n)!)) =1−(u^2 /2) +(u^4 /(4!))−.. ⇒1−(u^2 /2)≤cosu≤1−(u^2 /2)+(u^4 /(4!)) −1+(u^2 /2) −(u^4 /(4!))≤−cosu ≤−1+(u^2 /2) ⇒(u^2 /2)−(u^4 /(4!))≤1−cosu≤(u^2 /2) ⇒ (2/u^2 )≤(1/(1−cosu))≤(1/((u^2 /2)−(u^4 /(4!)))) ⇒ (1/(2x^2 ))≤(1/(1−cos(2x)))≤(1/(2x^2 −((16)/(4!))x^4 )) ⇒ ((e^x^2 −cosx)/(2x^2 ))≤((e^x^2 −cosx)/(1−cosx)) but lim_(x→+∞) ((e^x^2 −cosx)/(2x^2 )) =lim_(x→+∞) (e^x^2 /(2x^2 )) =+∞ ⇒ lim_(x→+∞) f(x)=+∞
$${let}\:{f}\left({x}\right)=\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{2}\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)} \\ $$$${we}\:{have}\:{cosu}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}−..\:\Rightarrow\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\leqslant{cosu}\leqslant\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!} \\ $$$$−\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\leqslant−{cosu}\:\leqslant−\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\leqslant\mathrm{1}−{cosu}\leqslant\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{2}}{{u}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}−{cosu}}\leqslant\frac{\mathrm{1}}{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}\leqslant\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{4}!}{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{2}{x}^{\mathrm{2}} }\leqslant\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{1}−{cosx}} \\ $$$${but}\:{lim}_{{x}\rightarrow+\infty} \:\:\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow+\infty} \frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} }\:=+\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$$$ \\ $$

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