Limit $\lim_{x \to \infty}\left (1- \frac1x\right)^x = ?$ $$\lim_{x \to \infty} \left(1+ \frac1x\right)^x = e$$
, 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.
 A: Try this:
$$
1-\frac{1}{x}=\frac{1}{\frac{x}{x-1}}=\frac{1}{1+\frac{1}{x-1}}
$$
so
$$
\lim_{x\rightarrow \infty} \left(1-\frac{1}{x}\right)^x=\lim_{x\rightarrow \infty} \left(\frac{1}{1+\frac{1}{x-1}}\right)^x=\lim_{x\rightarrow \infty} \frac{1}{\left(1+\frac{1}{x-1}\right)^{x-1}} \left(\frac{1}{1+\frac{1}{x-1}} \right)=\frac{1}{e}\cdot 1=\frac{1}{e}
$$
A: 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, $y=e^{-1}$.
A: $$\lim_{x \to \infty} (1- \frac1x)^x =e^{-1} $$ is evaluated  by taking logarithm of both sides of $ y=(1- \frac1x)^x $ and applying the L'Hospital Rule. 
A: 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,
$$\lim _{ x\rightarrow \infty  }{ (1+(-\frac { 1 }{ x } )^ x )}=e^(-1)$$
$$\lim _{ x\rightarrow \infty  }{ (1+(-\frac { 1 }{ x } )^ x )}=\frac1e$$
A: 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 $\frac{19}{20 e}\approx 0.349485$.
