Find $\lim\limits_{n\to \infty} [(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n}$ Find the value of 
$$\lim_{n\to \infty} \left[(1+\frac{1}{n})^n-(1+\frac{1}{n})\right]^{-n}$$.
I found the problem in one of my calculus books.I don't know it is asked before or not(If then command me I will check).I am struggling while doing this.
I started by taking logarithm.I take this logarithm to make the power to an algebraic form. But then the calculation became tough.I failed to do further. If I put the value of x in the expression I found the value is 0. But I need to find the exact value (i.e. the limiting). So please help me.
Thank you in advance!
 A: If $n\ge2$, then 
$$\left(1+{1\over n}\right)^n\ge1+{n\over n}+{n(n-1)\over2n^2}\gt{17\over8}$$ 
which means
$$\left(1+{1\over n}\right)^n-\left(1+{1\over n}\right)\gt{9\over 8}-{1\over n}$$
For $n\ge16$ (chosen to keep the arithmetic simple), we have
$$\left(1+{1\over n}\right)^n-\left(1+{1\over n}\right)\gt{17\over16}$$
and thus
$$\left(\left(1+{1\over n}\right)^n-\left(1+{1\over n}\right)\right)^{-n}\lt\left(16\over17\right)^n\to0$$
A: $$\lim_{n\to\infty} g(n) :=  \lim_{n\to\infty} \left (\left [1+\frac{1}{n}\right ]^n -\left [1+\frac{1}{n}\right ]\right )^{-n} = \lim_{n\to\infty} \left (1+\frac{1}{n}\right )^{-n}\left [\left (1+\frac{1}{n}\right )^{n-1}-1\right ]^{-n}$$
then $g(n)\xrightarrow[n\to\infty]{}0$, because the first term is bounded and the second term tends to $0$.
A: We have
\begin{align}
\lim_{n\to \infty} \left[\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n}\right)\right]^{-n} &= \lim_{n\to \infty} e^{-n\ln\left(\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n}\right)\right)} =0
\end{align}
because
$$
\underbrace{-n}_{\to -\infty}\cdot \underbrace{\ln\left(\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n}\right)\right)}_{\to \ln(e-1)} \to -\infty.
$$
A: Since $(1+\frac1n)^n\to e>2$ and $1+\frac1n\to1$, the term inside the parentheses is eventually greater than $1+\epsilon$ for some $\epsilon>0$. Thus your (positive) sequence is eventually bounded above by $(1+\epsilon)^{-n} \to0$. 
A: $$\begin{aligned}\left(\left(1+\frac1n\right)^n-\left(1+\frac1n\right)\right)^{-n}
&=\left(\left(1+x\right)^{1/x}-\left(1+x\right)\right)^{-1/x};\quad\quad\quad x=\frac1n
\\&=\left(\color{blue}{1}+\color{red}{\frac{1}{x}x}+\frac{(1/x)(1/x-1)}{2}x^2+\ldots+\binom{1/x}{i}x^i+\ldots-\left(\color{red}{1}+\color{blue}{x}\right)\right)^{-1/x};\quad |x|<1
\\&=\left(\color{blue}{1}-\color{blue}{x}+\frac{(1)(1-x)}{2}+\frac{(1)(1-x)(1-2x)}{3!}+\ldots+\left(\frac{1}{i!}\prod_{k=0}^{i-1}(1-kx)\right)\ldots\right)^{-1/x}
\\&=\exp\left[-\frac1x\ln\left(1-x+\frac{1-x}{2}+\ldots\right)\right]
\\&=\exp\left[-\frac{1}{x}\ln\left(e-1-\frac{e+2}2x+\mathcal{O}(x^2)\right)\right]
\end{aligned}$$
As $x\to0$, the term inside the logarithm approaches $(e-1)$ so the limit is $\lim_{x\to0}\exp\left[-\frac{\ln(e-1)}{x}\right]$, which is $0$.
For the coefficient of $x$, note that when expanding $\frac{(1-x)(1-2x)\ldots(1-(i-1)x)}{i!}$, we have $\frac{-x-2x-3x-\ldots}{i!}$, so the coefficient of $x$ is $-\left(1+\frac12+\frac{1+2}{3!}+\frac{1+2+3}{4!}+\ldots\right)=-\frac12\left[2+1+2\sum_{i=3}^{\infty}\frac{1}{i!}\frac{i(i-1)}{2}\right]=-\frac{\left(3+e-1\right)}2$.
A: We have that:
$$\lim_{h\to 0^+} (1+h)^{\frac{1}{h} } = \exp\left[\frac{\ln\left(1+h\right)}{h}\right] =e~~~~and ~~~\lim_{h\to 0^+} (1+h) = 1$$
Therefore Setting $\frac{1}{n}= h$
$$\lim_{n\to \infty} [(1+\frac{1}{n})^n-(1+\frac{1}{n})]^{-n} = \lim_{h\to 0} \exp\left[-\frac{\ln\left((1+h)^{\frac{1}{h}}-(1+h)\right)}{h}\right]\\=\lim_{h\to 0^+} \exp\left[-\frac{\ln\left((1+h)^{\frac{1}{h}}-(1+h)\right)}{h}\right]\\= \lim_{h\to 0^+} \exp\left[-\frac{1}{0^+}\cdot\ln\left(e-1\right)\right] = 0$$
Since $\ln(e-1)\sim \ln (1.7)>0$
