Limit

This limit can be written as

$\displaystyle \lim_{n \to \infty} \left(\ln(e^{n}) + \ln \left(\dfrac{n}{n + 1}\right)^{n^{2}}\right) = \lim_{n \to \infty} \left(n + n^{2} \ln \left(\dfrac{n}{n + 1}\right) \right) =$

$\displaystyle \lim_{n \to \infty} \left(n - n^{2} \ln \left(\dfrac{n + 1}{n}\right)\right) = \lim_{n \to \infty} \left(n - n^{2} \ln\left(1 + \dfrac{1}{n}\right) \right)$ .

Now the Maclaurin series for $\ln(1 + x)$ is

$\ln(1 + x) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \dfrac{x^{4}}{4} + ...$ for $-1 \lt x \le 1$ .

So with $x = \dfrac{1}{n}$ our limit becomes

$\displaystyle \lim_{n \to \infty} \left(n - n^{2}\left(\dfrac{1}{n} - \dfrac{1}{2n^{2}} + \dfrac{1}{3n^{3}} - \dfrac{1}{4n^{4}} + ... \right) \right) = \lim_{n \to \infty} \left(n - n + \dfrac{1}{2} - \dfrac{1}{3n} + \dfrac{1}{4n^{2}} - .... \right) =$

$\displaystyle \lim_{n \to \infty} \left(\dfrac{1}{2} + O\left(\dfrac{1}{n}\right) \right) = \dfrac{1}{2} = \boxed{0.5}$ .

Replacing $n$ by $\frac{1}{x}$ and further simplifying the expression, we get

$\large\lim_{x \rightarrow 0}$ [ $\frac{1}{x} -\frac{ln(x+1)}{x^2}$ ]

Now expanding,we get

$\frac{1}{x}- \frac{1}{x^2}[x-\frac {x^2}{2}+\frac{x^3}{3}-.............]$

$Ans.\boxed {\frac{1}{2}}$

The simpliest computation for limit $\begin{aligned}\lim_{n\to \infty} \ln\left[e^n\left(\frac{n}{n+1}\right)^{n^2}\right] &=\lim_{u \to \sqrt{e}}(\ln(u)) \quad{\text{Apply the chain rule: if } \lim_{u \to b}(f(u)) =L \space\ce{and} \lim_{u \to a}(g(x))=b \space\ce{and} f(x) \space\text{is continuous at } x=b. Then: \lim_{x \to a}(f(g(x))) =L }\\&= \ln(\sqrt{e}) \\&= \frac{1}{2} \space\space\square \end{aligned}$

$\large\text{FIN!!!}$