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Question Number 8633 by Yozzias last updated on 19/Oct/16

Find lim_(n→∞) (n^p /x^n )   where p∈Z^+ .

$$\mathrm{Find}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{n}^{\mathrm{p}} }{\mathrm{x}^{\mathrm{n}} }\:\:\:\mathrm{where}\:\mathrm{p}\in\mathbb{Z}^{+} . \\ $$

Commented by prakash jain last updated on 19/Oct/16

Does x cover entire range of R?  For ∣x∣≤1 limit is ∞.  For ∣x∣>1 limit is 0.  p≥1

$$\mathrm{Does}\:{x}\:\mathrm{cover}\:\mathrm{entire}\:\mathrm{range}\:\mathrm{of}\:\mathbb{R}? \\ $$$$\mathrm{For}\:\mid{x}\mid\leqslant\mathrm{1}\:\mathrm{limit}\:\mathrm{is}\:\infty. \\ $$$$\mathrm{For}\:\mid{x}\mid>\mathrm{1}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{0}. \\ $$$${p}\geqslant\mathrm{1} \\ $$

Commented by Yozzias last updated on 19/Oct/16

Please consider the ∣x∣>1 case! Thanks.

$$\mathrm{Please}\:\mathrm{consider}\:\mathrm{the}\:\mid\mathrm{x}\mid>\mathrm{1}\:\mathrm{case}!\:\mathrm{Thanks}. \\ $$

Answered by prakash jain last updated on 19/Oct/16

y=(n^p /x^n )  x>1  x^n =e^(nln x)   x^n =Σ_(i=0) ^∞ (((nln x)^i )/(i!))  (x^n /n^p )=Σ_(i=0) ^p (((ln x)^i )/(i!n^(p−i) ))+Σ_(i=p+1) ^∞ ((n^(i−p)  ln x)/(i!))  x>1⇒ln x>0  lim_(n→∞) (x^n /n^p )=0+∞=∞  or lim_(n→∞) (n^p /x^n )=0  for x<−1 case  or lim_(n→∞) (n^p /x^n )=lim_(n→∞) (1/((−1)^n ))(n^p /y^n )=0

$${y}=\frac{{n}^{{p}} }{{x}^{{n}} } \\ $$$${x}>\mathrm{1} \\ $$$${x}^{{n}} ={e}^{{n}\mathrm{ln}\:{x}} \\ $$$${x}^{{n}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}\mathrm{ln}\:{x}\right)^{{i}} }{{i}!} \\ $$$$\frac{{x}^{{n}} }{{n}^{{p}} }=\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\frac{\left(\mathrm{ln}\:{x}\right)^{{i}} }{{i}!{n}^{{p}−{i}} }+\underset{{i}={p}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{i}−{p}} \:\mathrm{ln}\:{x}}{{i}!} \\ $$$${x}>\mathrm{1}\Rightarrow\mathrm{ln}\:{x}>\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{{n}} }{{n}^{{p}} }=\mathrm{0}+\infty=\infty \\ $$$${or}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{p}} }{{x}^{{n}} }=\mathrm{0} \\ $$$${for}\:{x}<−\mathrm{1}\:{case} \\ $$$${or}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}^{{p}} }{{x}^{{n}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{{n}} }\frac{{n}^{{p}} }{{y}^{{n}} }=\mathrm{0} \\ $$$$ \\ $$

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