Límite de $e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$

Question

Que $c\ne0$ ser una constante. Considerar el límite de $$f(n)=e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$$ as $n # \to \infty$.

Creo que es cero porque % grande $n$,

$$e^{n^{3/4}} ((1- \frac{c}{n^{1/4}})^{n^{1/4}})^{\frac{n^{3/4}}{c}} \approx e^{n^{3/4}} (e^{-c})^{ \frac{n^{3/4}}{c}}$$

Pero, ¿cómo comprobarlo formalmente?

Answer 1

(24 / 10 / 2017) diversión downvote, puramente por razones matemáticas, estoy seguro de que...

Así, usando el % de ampliación $\log(1+x)=x-\frac12x^2+o(x^2)$cuando $x\to0$, uno ve que $$\log f(n)=n^{3/4}+c^{-1}n\log(1-cn^{-1/4})$$ is also $$\log f(n)=n^{3/4}+c^{-1}n\,(-cn^{-1/4}-\tfrac12c^2n^{-1/2}+o(n^{-1/2}))$$ that is, $$\log f(n)=-\tfrac12cn^{1/2}+o(n^{1/2})$$ In particular, $f # (n) \to0$ for every $c > 0 $ and $f (n) \to\infty$ for every $c < 0$.

Answer 2

Porque $\frac{-1}{1-t}\le-1-t$, $$\begin{align} \frac{\log(1-x)}x &=\frac1x\int_0^x\frac{-1}{1-t}\,\mathrm{d}t\\ &\le\frac1x\int_0^x(-1-t)\,\mathrm{d}t\\[3pt] &=-1-\frac{x}2 \end {alinee el} así $$ $$ \left(1-x\right) ^ {1 / x} \le e ^ {-1-x/2} $$ por lo tanto, $c\gt0$, $$\begin{align} e^{n^{3/4}}\left(\left(1-c/n^{1/4}\right)^{n^{1/4}}\right)^{n^{3/4}/c} &=e^{n^{3/4}}\left(\left(1-c/n^{1/4}\right)^{n^{1/4}/c}\right)^{n^{3/4}}\\ &\le e^{n^{3/4}}\left(e^{-1-\frac{c}{2n^{1/4}}}\right)^{n^{3/4}}\\[3pt] &=e^{-\frac{cn^{1/2}}2} \end {alinee el} $$

Answer 3

Si ponemos $ h=\frac{1}{n^{1/4}} $ luego

\begin{split}\lim_{n\to \infty}f(n)&=&\lim_{n\to \infty}e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}\\ &=& \lim_{n\to \infty}e^{n^{3/4}} (1- c/n^{1/4})^{n/c} \\&=&\lim_{h\to 0}e^{ \frac{1}{ h^3}}\left( 1-ch\right)^{1/ch^4} \\&=&\lim_{h\to 0}\exp\left( \frac{1}{ h^3}\right)\exp\left( \frac{\ln(1-ch)}{ch^4} \right)\\ &=&\lim_{h\to 0}\exp\left( \frac{1}{ h^3} \left( \frac{\ln(1-ch)}{ch}+1\right) \right) \\&=& \lim_{h\to 0}\exp\left( \frac{1}{ h^3} \left( -\frac{ch } {2} -\frac{c^2h^2}{3} -\frac{c^3h^3}{4} +o(h^3)\right) \right) \\&=&\lim_{h\to 0}\exp\left( -\frac{c } {2 h^2} -\frac{c^2}{3h} -\frac{c^3}{4} +o(1)\right)=0\end{dividido}

Dado que \ln(1-ch) $$ = -ch-\frac {c ^ 2 h ^ 2} {2}-\frac{c^3h^3}{3}-\frac{c^4h^4}{4} + o(h^4). $$