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$$\lim_{x \to \infty} \left(1+ \frac1x\right)^x = e$$ , then$$\lim_{x \to \infty} \left(1- \frac1x\right)^x =\; ?$$

I tried $\lim_{x \to \infty} \left(1- \frac1x\right)^x = \lim_{x \to \infty}\left(\frac{x-1}{x}\right)^x$, but this is not helpful.

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Try this:

$$ 1-\frac{1}{x}=\frac{1}{\frac{x}{x-1}}=\frac{1}{1+\frac{1}{x-1}} $$

so

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

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  • $\begingroup$ That is a lot more work than one needs. Have a look at the comment I left under the OP. $\endgroup$
    – Mark Viola
    May 22 '18 at 3:11
  • $\begingroup$ @MarkViola That is true, but I was just trying to give a different approach that perhaps used previously established facts. I understand why your last inequality holds, but perhaps the OP didn't $\endgroup$
    – JasonM
    May 22 '18 at 4:13
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Let, $$y=(1-\frac{1}{x})^x$$

$$or, \ln y=x\ln(1-\frac{1}{x})$$ $$or,\ln y=\frac{\ln(1-\frac{1}{x})}{x^{-1}}$$

Now, $$or,\displaystyle\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{1}{x})}{x^{-1}}$$

Using L'Hospital rule, $$\displaystyle\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}-\frac{1}{1-\frac{1}{x}}=-1$$ So, $y=e^{-1}$.

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  • $\begingroup$ The use of logarithms is completely unnecessary. $\endgroup$
    – Mark Viola
    May 22 '18 at 3:11
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$$\lim_{x \to \infty} (1- \frac1x)^x =e^{-1} $$ is evaluated by taking logarithm of both sides of $ y=(1- \frac1x)^x $ and applying the L'Hospital Rule.

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Apply the common limit:$\lim _{ x\rightarrow \infty }{ (1+\frac { k }{ x } )^ x } =e^ k$

In this case we have,$$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}$$ So we compare it with $\lim _{ x\rightarrow \infty }{ (1+\frac { k }{ x } )^ x } =e^ k$ we have, $k=-1$ and $x=x$

So, $$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}=e^(-1)$$ $$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}=\frac1e$$

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Just another way.

$$y=\left(1-\frac{1}{x}\right)^x\implies \log(y)=x \log\left(1-\frac{1}{x}\right)$$ Now, using Taylor series, $$\log\left(1-\frac{1}{x}\right)=-\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\log(y)=-1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ Continuing with Taylor series $$y=e^{\log(y)}=\frac{1}{e}-\frac{1}{2 e x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.

Even for "rather small" values of $x$, the approximation is quite good. For example, for $x=10$, the exact value is $3486784401 \times 10^{-10}=0.3486784401$ while the above formula gives $\frac{19}{20 e}\approx 0.349485$.

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