Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)=+∞

$${let}\:{f}\left({x}\right)=\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{2}\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)} \\ $$$${we}\:{have}\:{cosu}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}−..\:\Rightarrow\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\leqslant{cosu}\leqslant\mathrm{1}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{4}} }{\mathrm{4}!} \\ $$$$−\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\leqslant−{cosu}\:\leqslant−\mathrm{1}+\frac{{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}\leqslant\mathrm{1}−{cosu}\leqslant\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{2}}{{u}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}−{cosu}}\leqslant\frac{\mathrm{1}}{\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}!}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}\leqslant\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{16}}{\mathrm{4}!}{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{2}{x}^{\mathrm{2}} }\leqslant\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{1}−{cosx}} \\ $$$${but}\:{lim}_{{x}\rightarrow+\infty} \:\:\frac{{e}^{{x}^{\mathrm{2}} } −{cosx}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$={lim}_{{x}\rightarrow+\infty} \frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} }\:=+\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com