Is $\lim \limits_{n\rightarrow \infty} n^{1-\ln((1+\frac{1}{n})^n)} = 1 $ could you help me understand if this statement is correct?
$$ \lim \limits_{n\rightarrow \infty} n^{1-\ln((1+\frac{1}{n})^n)} = 1 $$
Its easy to see that $ \lim \limits_{n\rightarrow \infty} 1-\ln((1+\frac{1}{n})^n) = 0 $, but since this expression is the exponent of $n$, I don't know if you can conclude that 
$$ \lim \limits_{n\rightarrow \infty} n^{1-\ln((1+\frac{1}{n})^n)} = \lim \limits_{n\rightarrow \infty} n^0 =1 $$
 A: The esteemed Kavi Rama Murthy has explained why your attempted proof is not correct.
To show the result, it suffices to show that the logarithm of that expression tends to zero, i.e.
$$\left(1 - n \ln (1 + 1/n)\right) \ln n \to 0.$$
Note that by Taylor's theorem, $|\ln(1+1/n) - \frac{1}{n}| \le \frac{C}{n^2}$ for some constant $C>0$.
Thus,
$$|1 - n \ln(1+1/n)| (\ln n)
\le \frac{C}{n} \ln n \to 0.$$

Response to comment: Using the mean value form of the remainder, the remainder for the first-order Taylor polynomial is $\ln(1+x) - x = - \frac{1}{(1+\xi_x)^2} \frac{x^2}{2}$
where $\xi_x$ is a quantity dependent on $x$ that is between $x$ and $0$. When $x$ is between $0$ and $1$, we have $|\frac{1}{(1+\xi_x)^2}| \le 1$.
A: You have an indeterminate form of the type $\infty^0$. You can use the fact that for a continuous function $g$ you have that $\displaystyle g\left(\lim_{x\to a}f(x)\right)=\lim_{x\to a}g(f(x))$. In this case you can take $g= \ln$ and use angryavian's argument to show that the logarithm of the expression tends to zero.
