Question on limits containing logarithm

Question

Let $a=n^ {(n+\frac{1} { ln(n)})}$ and $b=(n+\frac{1}{ln(n)})^n$ . The value of $$\lim\limits_{n \to \infty} \frac{a}{b}$$

My Approach : I started this problem by first taking log on both sides of $a $ and $b$ but it did not become someting logical ,I also used some approximations but did not reached to any conclusion.

Please help ...

Answer 1

$$\begin{align*}&\log a=\left(n+\frac1{\log n}\right)\log n=n\log n+1\\{}\\&\log b=n\log\left(n+\frac1{\log n}\right)\end{align*}$$

and now using some Taylor series:

$$n\log\left(n+\frac1{\log n}\right)=n\log n+n\log\left(1+\frac1{n\log n}\right)=$$

$$=n\log n+n\left(\frac1{n\log n}-\frac1{2n^2\log^2 n}+\ldots\right)=n\log n+\frac1{\log n}+\mathcal O\left(\frac1{n\log n}\right)$$

Thus:

$$\log\frac ab=\log a-\log b=1-\frac1{\log n}+\mathcal O\left(\frac1{n\log n}\right)\xrightarrow[n\to\infty]{}1$$

so that

$$\frac ab=e^{\log\frac ab}\xrightarrow[n\to\infty]{}e$$

Answer 2

This avoids Taylor's series expansions. Let $L$ be the desired limit then we have \begin{align} \log L &= \log\left\{\lim_{n \to \infty}\frac{a}{b}\right\}\notag\\ &= \lim_{n \to \infty}\log (a/b)\text{ (via continuity of log)}\notag\\ &= \lim_{n \to \infty}\log a - \log b\notag\\ &= \lim_{n \to \infty}\left(n + \frac{1}{\log n}\right)\log n - n\log\left(n + \frac{1}{\log n}\right)\notag\\ &= 1 - \lim_{n \to \infty}n\log\left(\dfrac{n + \dfrac{1}{\log n}}{n}\right)\notag\\ &= 1 - \lim_{n \to \infty}n\log\left(1 + \dfrac{1}{n\log n}\right)\notag\\ &= 1 - \lim_{n \to \infty}n\cdot\dfrac{1}{n\log n}\cdot\dfrac{\log\left(1 + \dfrac{1}{n\log n}\right)}{\dfrac{1}{n\log n}}\notag\\ &= 1 - \lim_{n \to \infty}\frac{1}{\log n}\notag\\ &= 1\notag \end{align} and hence $L = e$. Here we have used the fact that $\log n \to \infty$ as $n \to \infty$ and $(1/x)\log(1 + x) \to 1$ as $x \to 0$.