Menu Close

Calculus-lim-x-x-2-2-x-2-x-x-x-x-x-

Tinku Tara 0 Comments
Pin




Question Number 138888 by mathdanisur last updated on 19/Apr/21
Calculus: lim_(x→∞) [ (x^2 /2^x ) + (2^x /(x!)) + ((x!)/x^x ) ] = ?
$${Calculus}: \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\:\left[\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{{x}} }\:+\:\frac{\mathrm{2}^{{x}} }{{x}!}\:+\:\frac{{x}!}{{x}^{{x}} }\:\right]\:=\:? \\ $$
Answered by mathmax by abdo last updated on 20/Apr/21
lim_(x→+∞) (x^2 /2^x )=lim_(x→+∞) x^2 2^(−x) =0 x! ∼ x^x e^(−x) (√(2πx))( +∞) ⇒(2^x /(x!))∼(2^x /(x^x e^(−x) (√(2πx)))) =((2^x x^(−x) e^x )/( (√(2πx)))) =((e^(xlog2) .e^(−xlogx) e^x )/( (√(2πx)))) =(e^(xlog2−xlogx+x) /( (√(2πx)))) =(e^(x(log2−logx +1)) /( (√(2πx)))) →0(x→+∞) ((x!)/x^x )∼e^(−x) (√(2πx))→0 ⇒lim_(x→+∞) ((x^2 /2^x )+(2^x /(x!)) +((x!)/x^x ))=0
$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{x}} }=\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{x}^{\mathrm{2}} \:\mathrm{2}^{−\mathrm{x}} \:=\mathrm{0}\:\:\: \\ $$$$\mathrm{x}!\:\sim\:\mathrm{x}^{\mathrm{x}} \mathrm{e}^{−\mathrm{x}} \sqrt{\mathrm{2}\pi\mathrm{x}}\left(\:+\infty\right)\:\Rightarrow\frac{\mathrm{2}^{\mathrm{x}} }{\mathrm{x}!}\sim\frac{\mathrm{2}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{x}} \:\mathrm{e}^{−\mathrm{x}} \sqrt{\mathrm{2}\pi\mathrm{x}}}\:=\frac{\mathrm{2}^{\mathrm{x}} \:\mathrm{x}^{−\mathrm{x}} \:\mathrm{e}^{\mathrm{x}} }{\:\sqrt{\mathrm{2}\pi\mathrm{x}}} \\ $$$$=\frac{\mathrm{e}^{\mathrm{xlog2}} \:.\mathrm{e}^{−\mathrm{xlogx}} \:\mathrm{e}^{\mathrm{x}} }{\:\sqrt{\mathrm{2}\pi\mathrm{x}}}\:=\frac{\mathrm{e}^{\mathrm{xlog2}−\mathrm{xlogx}+\mathrm{x}} }{\:\sqrt{\mathrm{2}\pi\mathrm{x}}}\:=\frac{\mathrm{e}^{\mathrm{x}\left(\mathrm{log2}−\mathrm{logx}\:+\mathrm{1}\right)} }{\:\sqrt{\mathrm{2}\pi\mathrm{x}}}\:\rightarrow\mathrm{0}\left(\mathrm{x}\rightarrow+\infty\right) \\ $$$$\frac{\mathrm{x}!}{\mathrm{x}^{\mathrm{x}} }\sim\mathrm{e}^{−\mathrm{x}} \sqrt{\mathrm{2}\pi\mathrm{x}}\rightarrow\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{x}} }+\frac{\mathrm{2}^{\mathrm{x}} }{\mathrm{x}!}\:+\frac{\mathrm{x}!}{\mathrm{x}^{\mathrm{x}} }\right)=\mathrm{0} \\ $$
Commented by mathdanisur last updated on 20/Apr/21
thank you very much sir
$${thank}\:{you}\:{very}\:{much}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 20/Apr/21
lim_(x→+∞) x^2 2^(−x) =∞.0 ( indeterminate form)≠0 Please sir explain.
$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{x}^{\mathrm{2}} \:\mathrm{2}^{−\mathrm{x}} =\infty.\mathrm{0}\:\left(\:{indeterminate}\:{form}\right)\neq\mathrm{0} \\ $$$${Please}\:\boldsymbol{{sir}}\:{explain}. \\ $$
Commented by mnjuly1970 last updated on 20/Apr/21
lim_(x→∞) (x^2 /2^x ) =_(hopital) ^(∞/∞) lim_(x→+∞) ((2x)/(2^x .ln(x))) =lim_(x→∞) (2/(2^x .ln^2 (x)))=((→2)/(→∞)) →0
$$\:\:{lim}_{{x}\rightarrow\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{2}^{{x}} }\:\underset{{hopital}} {\overset{\frac{\infty}{\infty}} {=}}\:{lim}_{{x}\rightarrow+\infty} \frac{\mathrm{2}{x}}{\mathrm{2}^{{x}} .{ln}\left({x}\right)} \\ $$$$\:\:\:\:={lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}}{\mathrm{2}^{{x}} .{ln}^{\mathrm{2}} \left({x}\right)}=\frac{\rightarrow\mathrm{2}}{\rightarrow\infty}\:\rightarrow\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 20/Apr/21
But mathmax sir didn′t use l′hopital rule.He evaluated the limit directly! Anyway you′re rivht and thanks sir!
$$\mathcal{B}{ut}\:{mathmax}\:{sir}\:{didn}'{t}\:{use}\:{l}'{hopital} \\ $$$${rule}.{He}\:{evaluated}\:{the}\:{limit}\:{directly}! \\ $$$$\mathcal{A}{nyway}\:{you}'{re}\:{rivht}\:{and}\:{thanks}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 20/Apr/21
Thαnks very much sir!
$$\mathcal{T}{h}\alpha{nks}\:{very}\:{much}\:{sir}! \\ $$
Commented by mathmax by abdo last updated on 20/Apr/21
i use that e^x defeat all polynom at ∞...!
$$\mathrm{i}\:\mathrm{use}\:\mathrm{that}\:\mathrm{e}^{\mathrm{x}} \:\mathrm{defeat}\:\mathrm{all}\:\mathrm{polynom}\:\mathrm{at}\:\infty…! \\ $$
Commented by mathmax by abdo last updated on 20/Apr/21
let prove this result f(x)=x^2 2^(−x) =x^2 e^(−xlog2) changement e^(−xlog2) =t(t→0) ⇒−xlog2=logt ⇒x=−((logt)/(log2)) ⇒ x^2 e^(−xlog2) =−((log^2 t)/(log^2 2))×t =−((tlog^2 t)/(log^2 2)) due to lim_(t→0) tlogt=0 ⇒ lim_(x→+∞) x^2 e^(−xlog2) =0
$$\mathrm{let}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{result}\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \:\mathrm{2}^{−\mathrm{x}} \:=\mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{xlog2}} \\ $$$$\mathrm{changement}\:\mathrm{e}^{−\mathrm{xlog2}} =\mathrm{t}\left(\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow−\mathrm{xlog2}=\mathrm{logt}\:\Rightarrow\mathrm{x}=−\frac{\mathrm{logt}}{\mathrm{log2}}\:\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} \:\mathrm{e}^{−\mathrm{xlog2}} =−\frac{\mathrm{log}^{\mathrm{2}} \mathrm{t}}{\mathrm{log}^{\mathrm{2}} \mathrm{2}}×\mathrm{t}\:=−\frac{\mathrm{tlog}^{\mathrm{2}} \mathrm{t}}{\mathrm{log}^{\mathrm{2}} \mathrm{2}}\:\:\mathrm{due}\:\mathrm{to}\:\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{tlogt}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{xlog2}} \:=\mathrm{0} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *