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A bit rusty on my calculus, trying to figure out whether this is correct.

I tried proving $\underset{x \rightarrow \infty}{\lim} (1-e^{-x})^{x} = 1$ without using L'hôpital.

$$\lim_{x \rightarrow \infty} \left(1-e^{-x}\right)^{x}=\lim_{x \rightarrow \infty}\left(1- \frac{x\cdot e^{-x}}{x}\right)^{x}\overset{(1)}{=}\lim_{x \rightarrow \infty} e^{e^{-x}\cdot x}=e^{\lim_{x \rightarrow \infty} e^{\frac{x}{e^{x}}}}=e^{0}=1$$

I'm a little unsure about the transition I marked as $(1)$.

Arctic Char
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ec2604
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  • Taking $\log(\cdot)$, it suffices to show $$\lim_{x\to\infty}x \log(1 - e^{-x}) = 0. $$ However, by the L'Hopital's rule, $$\lim_{x\to\infty}x \log(1 - e^{-x}) = \lim_{x\to\infty}\frac{\log(1 - e^{-x})}{1/x} = \lim_{x\to\infty}\frac{e^{-x}/(1-e^{-x})}{-1/x^2} = \lim_{x\to\infty}\frac{(-x^2)e^{-x}}{1-e^{-x}}=0.$$ – Sangchul Lee Nov 10 '20 at 13:42
  • Please avoid $\rm\LaTeX$ only titles (also @Varun who edited the title). – Asaf Karagila Nov 10 '20 at 13:43
  • @SangchulLee the OP did not want to use H's rule... – peter a g Nov 10 '20 at 13:48
  • @AsafKaragila Thanks for pointing this out to me, I'll keep that in mind from now on. – Varun Vejalla Nov 10 '20 at 13:51
  • @peterag Thank you, I did mean $x \to \infty$, so let me repeat: $\lim_{x \to \infty} \left( 1-\frac nx \right)^x = e^{-n}$ for a constant $n$, but if $n$ depends on $x$ the plot thickens. – Sarvesh Ravichandran Iyer Nov 10 '20 at 13:55

1 Answers1

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We have that as $x \to \infty$

$$\left(1+\frac a{x}\right)^{x} \to e^a$$

is true only for $a\in \mathbb R$ fixed, here we can use that

$$\left(1-\frac1{e^{x}}\right)^{x}=e^{x\log\left(1-\frac1{e^{x}}\right) }=e^{\frac{x}{e^x}\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} }$$

and use standard limits to get the result indeed

  • $\frac{x}{e^x} \to 0$
  • $\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} \to -1$

Refer also to

user
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  • Doesn't this mean I need to use l'hopital? – ec2604 Nov 10 '20 at 13:43
  • @ec2604 No we don't need l'Hospital at all. I'll give some reference for that. – user Nov 10 '20 at 13:45
  • OK. Not quite sure how you get the second limit without l'Hôpital. It's an undetermined form so it seems like the obvious thing to do. – ec2604 Nov 10 '20 at 13:47
  • @ec2604 We have that as $t \to 0$ by standard limit $t\log(1+t)=\log [(1+t)^t] \to \log e=1$ – user Nov 10 '20 at 13:49
  • Or $$\lim_{x\to0}\frac{\log(1-x)}x=-\lim_{x\to0}\frac{\log(1-x)-\log(1-0}{-x-0)}$$ and the last expression is a derivative. – saulspatz Nov 10 '20 at 13:52
  • @saulspatz Yes of course, even if often the derivative is proved by the fundamental limit $(1+1/x)^x \to e$. – user Nov 10 '20 at 13:54