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Let $x>0$ be a constant. Show that $\lim_{z \rightarrow \infty} -z^xe^{-z} = 0$.

Do I apply L'hopital's rule here? It doesn't seem to help, for example, after one round of differentiating the numerator and denominator, I get $\frac{d}{dz} z^x = xz^{x-1}$ and $\frac{d}{dz}e^z = e^z$. I am unsure on how to proceed.

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  • $\begingroup$ Is it $(-z)^x$ or $-(z^x)$? $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 12:34
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    $\begingroup$ Hint:$$z^xe^{-z}=x^x\left(\frac zxe^{-\frac zx}\right)^x$$Do you know how to evaluate$$\lim_{t\to\infty}te^{-t}$$where $t=z/x$? $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 12:36
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    $\begingroup$ @pasabaporaqui Suppose $x=1/2$. How can you apply L'H half of a time? $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 12:38
  • $\begingroup$ @SimplyBeautifulArt Thanks, your hint is really elegant! Could you post it as an answer so I can accept it? $\endgroup$ – elbarto Sep 10 '17 at 12:57
  • $\begingroup$ Sure =) @elbarto $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 12:58
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Start by showing with L'Hospital's that

$$\lim_{t\to\infty}\frac t{e^t}=\lim_{t\to\infty}\frac1{e^t}=0$$

Now for any $x>0$, set $t=z/x$ to get

$$\lim_{z\to\infty}\frac{z^x}{e^z}=\left[\lim_{z\to\infty}\frac z{e^{z/x}}\right]^x=\left[x\lim_{z\to\infty}\frac t{e^t}\right]^x=0$$

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let $m\in \mathbb{Z}$ and $m\ge x$ then $$0<\frac { { z }^{ x } }{ { e }^{ z } } \le { \frac { { z }^{ m } }{ { e }^{ z } } = }{ \left( \frac { z }{ \sqrt [ m ]{ { { e }^{ z } } } } \right) }^{ m }={ \left( \frac { z }{ { b }^{ z } } \right) }^{ m }$$ where $b=\sqrt [ m ]{ e } >1$ Then we have $$\\ 0<\frac { { z } }{ { b }^{ z } } =\frac { z }{ { \left( 1+\left( b-1 \right) \right) }^{ z } } =\frac { z }{ 1+z\left( b-1 \right) +\frac { z\left( z-1 \right) }{ 2 } { \left( b-1 \right) }^{ 2 }+...+{ \left( b-1 \right) }^{ z } } <\frac { 2z }{ z\left( z-1 \right) { \left( b-1 \right) }^{ 2 } } \overset { z\rightarrow \infty }{ \longrightarrow 0 } \\ \\ $$ so we get $${ \left( \frac { z }{ { b }^{ z } } \right) }^{ m }\rightarrow 0$$,when $z\rightarrow \infty $

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    $\begingroup$ the statement says "show that.." so I showed $\endgroup$ – haqnatural Sep 10 '17 at 12:59
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    $\begingroup$ @pasabaporaqui If you are so concerned with L'H, please post your own answer instead of commenting on other answers... $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:00
  • $\begingroup$ @pasabaporaqui and that gives you no reason whatsoever to post L'H comments on non-L'H answers. $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:16
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let $$L=\lim_{z\to \infty} z^xe^{-z}$$ Taking log $$\ln L=\lim_{z\to \infty} -xz\ln z$$ Now $$\lim_{z\to \infty} z\ln z =\infty$$

Thus \begin{align}\ln L=-x\infty\end{align} If $x\gt 0$ \begin{align}\ln L=-\infty\\ L=e^{-\infty}\\L=0\end{align}

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  • $\begingroup$ If you would mind my editing the spoiler for you? $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:04
  • $\begingroup$ =) Hope that's what you had intended $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:06
  • $\begingroup$ Yes it would be great $\endgroup$ – neonpokharkar Sep 10 '17 at 13:06
  • $\begingroup$ I believe you meant to have $z\to\infty$. $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:07
  • $\begingroup$ Yes, and how do you do it? $\endgroup$ – neonpokharkar Sep 10 '17 at 13:08
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i replace z with x(variable) and x with a(parameter) then we have $\lim_{x \rightarrow \infty} -x^ae^{-x} $

$\displaystyle x^ae^{-x}= \frac{e^{a\ln x}}{e^x}=e^{a\ln x-x}$

$\displaystyle \lim_{x \to \infty} (a\ln x-x)=- \infty$

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  • $\begingroup$ the second limit is not complex $\endgroup$ – Lerigorilla Sep 10 '17 at 13:00
  • $\begingroup$ the derivative of alnx tends to zero while -x in -1 so it goes in infinity $\endgroup$ – Lerigorilla Sep 10 '17 at 13:02
  • $\begingroup$ so the derivative of alnx-x tends to -1 and we are done $\endgroup$ – Lerigorilla Sep 10 '17 at 13:06

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