$$ \lim_{x\to \infty} ({1-\frac{1}{x}})^x $$
I can solve it using L'Hopital, but not without. No taylor series just algebra
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please solve this without taylor's series or whatever other things are called, just use algebra – Mouse Hello Oct 12 '16 at 15:32
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1It is answered already in this question. – Dietrich Burde Oct 12 '16 at 15:39
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@Jean-ClaudeArbaut i mean compute it without derivatives or any other fancy techniques – Mouse Hello Oct 12 '16 at 15:41
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Are you allowed at least to use the limit $(1+1/n)^n\to e$, as answers below do? – Jean-Claude Arbaut Oct 12 '16 at 15:42
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yes @Jean-ClaudeArbaut – Mouse Hello Oct 12 '16 at 15:44
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$$\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^x=\lim_{x\to\infty}\left(\frac{x-1}{x}\right)^x=\lim_{x\to\infty}\left(\frac{x}{x-1}\right)^{-x}=\lim_{x\to\infty}\left(1+\frac{1}{x-1}\right)^{-x}=$$ $$=\left\{u=x-1\right\}=\lim_{u\to\infty}\left(1+\frac{1}{u}\right)^{-\left(u+1\right)}=\lim_{u\to\infty}\left(1+\frac{1}{u}\right)^{-u}\cdot \lim_{u\to\infty}\left(1+\frac{1}{u}\right)^{-1}=\frac{1}{e}\cdot 1=\frac{1}{e}$$
Note that splitting the limit to a product of limits is fine since both limits exist.
Guy
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1These are the secrets of algebraic manipulations. Guy just got a white bunny from his hat :) – imranfat Oct 12 '16 at 15:46
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u = x-1 -> -(u+1) = -x =/= -u ? what do you mean? @Jean-ClaudeArbaut – Mouse Hello Oct 12 '16 at 15:55
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1@MouseHello I mean $-(u+1)$ has not become $-u$ and $-1$ has magically disappeared, but $-1$ is still here and you don't look at it. – Jean-Claude Arbaut Oct 12 '16 at 15:56
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Thanks so much for the help @Jean-ClaudeArbaut i've already understood it now. quick question: how can guy just think up the 4th step so quickly? that's pretty cool – Mouse Hello Oct 12 '16 at 16:03
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$$ \lim_{x\to \infty}\left({1-\frac{1}{x}}\right)^x=\lim_{x\to \infty} \frac1{\left(1+\dfrac 1{x-1}\right)^{x}}.$$
Then you can replace $x-1$ by $x$ without changing the result and the limit is $\dfrac1e$.