How to calculate this limit expression? I am having a difficult time with the limit:
$$\lim_{n\to \infty} \left(1+\log\left(\frac{n}{n-1}\right)\right)^n $$
How should I approach this problem?
 A: Herein, we present a way forward that does not rely on differential calculus, but rather uses an elementary pair of inequalities and the squeeze theorem.  To that end we proceed.

PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag1$$
for $x>0$.

Using $(1)$ with $x=\frac{n}{n-1}$, we find that
$$\frac{n+1}{n}\le 1+\log\left(\frac{n}{n-1}\right)\le \frac{n}{n-1} \tag 2$$
Finally, from $(2)$ we find that
$$\left(1+\frac1n\right)^n\le \left(1+\log\left(\frac{n}{n-1}\right)\right)^n\le \frac{1}{\left(1-\frac1n\right)^n} \tag 3$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to  \infty} \left(1+\log\left(\frac{n}{n-1}\right)\right)^n=e}$$
A: $$\lim _{n\to \infty }\:\left(1+\log \left(\frac{n}{n-1}\right)\right)^n\: = \lim _{n\to \infty }\:\left(e^{n\cdot \:ln\left(1+\log \left(\frac{n}{n-1}\right)\right)}\right)\: \approx$$
$$\lim _{n\to \infty }\:\left(e^{n\cdot \:ln\left(\frac{n}{n-1}\right)}\right)\: \approx \lim _{n\to \infty }\:\left(e^{n\cdot \left(\frac{n-n+1}{n-1}\right)}\right)\: = \color{red}{e}$$
Note: $ln(\frac{n}{n-1})$ if the argumenent of the ln $\frac{n}{n-1} \to 1$ for $n \to \infty$ then $$\color{red}{ln\left(\frac{n}{n-1}\right) \approx \left(\frac{n-\left(denominator\right)}{n-1}\right) = \left(\frac{n-\left(n-1\right)}{n-1}\right) = \left(\frac{1}{n-1}\right)}$$
A: Let the Limit be equal to $L$. Then $\lim n(1+ \log(n/(n-1))=\ln L$. This implies: 
$$\lim\frac{1+\log(1/(1-1/n))}{1/n}$$ Let $y=1/n$.$$\lim\frac{1-\log(1-y)}{y}=\lim \frac{1/(1-y)}{1}=1$$ But $1=\ln L$ which implies $L=e$.
A: By applying an exponential of a logarithm to the expression we get
$$\lim_{x\to\infty} \left( 1+\log \left( \frac{x}{x-1} \right)\right)^x=\lim_{x\to\infty} \mathrm{exp}\left( x\log \left( 1+\log \left( \frac{x}{x-1}\right)\right)\right)$$
We will show that $$L:=\lim_{x\to\infty} \left( x\log \left( 1 + \log \left( \frac{x}{x-1}\right) \right) \right)=1$$ and thus the result is $\lim_{x\to\infty} \mathrm{exp} (1)=\color{red} e$.
$$L= \lim_{x\to\infty} \left( x\log \left( 1 + \log \left( \frac{x}{x-1}\right) \right) \right)=\lim_{x\to\infty} \left(\dfrac{\log \left( 1+\log\left( \frac{x}{x-1}\right)\right)}{\dfrac 1x}\right)$$
so by de l'Hospital's rule
$$L=\lim_{x\to\infty} \left( \dfrac{-\dfrac{1}{x(x-1)\left( \log \left(\frac{x}{x-1} \right) + 1\right)}}{-\dfrac{1}{x^2}}\right)=\lim_{x\to\infty} \left( \dfrac{x}{(x-1)\left( \log \left( \frac{x}{x-1}\right) + 1\right)}\right)$$
Use de l'Hospital's rule again to get
$$L=\lim_{x\to\infty} \left( \dfrac{1}{\log \left( \frac{x}{x-1}\right) +1 - \frac 1x}\right)=\dfrac{\lim_{x\to\infty} 1}{\lim_{x\to\infty} \left( \log \left( \frac{x}{x-1}\right) +1 - \frac 1x\right)}=\frac 1 1=1$$
as desired.
