Limit of $ \frac1{n -\log n}$ as $n$ approaches $\infty$.

Question

I am not able to find the following limit. $$\lim_{n\to \infty} \frac{1}{n-\log n}$$

I tried replacing log function with it's expansion but soon stuck. Also tried dividing both numerator & denominator by $n$ to get the following $$\lim_{n\to \infty} \frac{\frac{1}{n}}{1-\frac{\log\ n}{n}}$$ but couldn't proceed further. Can I break the numerator & denominator into $2$ separate limits ? Please also suggest how to calculate this limit? (You can replace $n$ by $n+1$ here)

Answer 1

Simply note that:

$$\frac{1}{n-\log n}=\frac{1}{n}\frac{1}{1-\frac{\log n}{n}}\to 0\cdot 1=0$$

Answer 2

Guide:

Note that we have $\lim_{n \to \infty} \frac{\log n}{n} = 0$, you can prove that using L'hopital's rule and you can use that to answer your question.

Answer 3

Using equivalents for large $n$ (remembering that $\log(n)< n)$ $$\frac{1}{n-\log (n)}=\frac{1}{n}\frac{1}{1-\frac{\log (n)}{n}}\sim \frac{1}{n}\left(1+\frac{\log (n)}{n} \right)=\frac{1}{n}+\frac{\log (n)}{n^2}$$