Question Number 162399 by cortano last updated on 29/Dec/21
$$\:\:{Let}\:{m}\:\&\:{n}\:{be}\:{two}\:{positive}\:{numbers}\: \\ $$$$\:{greater}\:{than}\:\mathrm{1}\:.\:{If}\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{cos}\:\left({p}^{{n}} \right)} −{e}}{{p}^{{m}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}{e}\: \\ $$$$\:{then}\:\frac{{n}}{{m}}=? \\ $$
Commented by blackmamba last updated on 29/Dec/21
![lim_(p→0) ((e^(cos (p^n )) −e)/p^m ) = e lim_(p→0) ((e^(cos (p^n )−1) −1)/p^m ) = (1/2)e lim_(p→0) ((e^(cos (p^n )−1) −1)/(cos (p^n )−1)) ×lim_(p→0) ((cos (p^n )−1)/p^m ) = (1/2) [ notice that lim_(x→0) ((e^x −1)/x) = 1 ] [ lim_(p→0) ((e^(cos (p^n )−1) −1)/(cos (p^n )−1)) = 1 ] ⇔ lim_(p→0) ((cos (p^n )−1)/p^m ) = (1/2) ⇔ lim_(p→0) ((−2sin^2 ((p^n /2)))/p^m ) =(1/2) ⇔ lim_(p→0) (((((sin ((p^n /2)))/(((p^n /2)))))^2 .((p^(2n) /4)))/p^m ) =^? −(1/4) 2n=m ⇒(n/m)= (n/(2m)) = (1/2)](https://www.tinkutara.com/question/Q162421.png)
$$\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{cos}\:\left({p}^{{n}} \right)} −{e}}{{p}^{{m}} }\:=\:{e}\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}} −\mathrm{1}}{{p}^{{m}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}{e} \\ $$$$\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}} −\mathrm{1}}{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}}\:×\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}}{{p}^{{m}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\left[\:{notice}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}\:=\:\mathrm{1}\:\right] \\ $$$$\:\left[\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}} −\mathrm{1}}{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}}\:=\:\mathrm{1}\:\right]\: \\ $$$$\:\Leftrightarrow\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left({p}^{{n}} \right)−\mathrm{1}}{{p}^{{m}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\Leftrightarrow\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{{p}^{{n}} }{\mathrm{2}}\right)}{{p}^{{m}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\Leftrightarrow\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{sin}\:\left(\frac{{p}^{{n}} }{\mathrm{2}}\right)}{\left(\frac{{p}^{{n}} }{\mathrm{2}}\right)}\right)^{\mathrm{2}} .\left(\frac{{p}^{\mathrm{2}{n}} }{\mathrm{4}}\right)}{{p}^{{m}} }\:\overset{?} {=}\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\mathrm{2}{n}={m}\:\Rightarrow\frac{{n}}{{m}}=\:\frac{{n}}{\mathrm{2}{m}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by qaz last updated on 29/Dec/21
$$\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{cos}\:\left(\mathrm{p}^{\mathrm{n}} \right)} −\mathrm{e}}{\mathrm{p}^{\mathrm{m}} } \\ $$$$=\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}\left(\mathrm{e}^{\mathrm{cos}\:\left(\mathrm{p}^{\mathrm{n}} \right)−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{p}^{\mathrm{m}} } \\ $$$$=\mathrm{e}\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{p}^{\mathrm{n}} \right)−\mathrm{1}}{\mathrm{p}^{\mathrm{m}} } \\ $$$$=\mathrm{e}\underset{\mathrm{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{p}^{\mathrm{2n}} }{\mathrm{p}^{\mathrm{m}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e} \\ $$$$\Rightarrow\frac{\mathrm{n}}{\mathrm{m}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$