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

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)

• If you are ok, you can set as solved. Thanks! – user Dec 15 '17 at 21:42

## 3 Answers

Simply note that:

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

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.

• But that would require separating limits of numerator & denominator & considering them as 2 separate functions. Can we do that? – Anuj Dec 12 '17 at 23:35
• check here point $4$. check the limit of the denominator is non-zero. – Siong Thye Goh Dec 12 '17 at 23:36
• thanks ! Just 1 more clarification reqd. Assume a case, where I have to find limit of f(x)*g(x) .If separately calculated, limit of f(x) comes out as infinity & that of g(x) as 4. Can we say f(x)*g(x) tends to infinity – Anuj Dec 12 '17 at 23:42
• yes, when $x$ is huge, $f(x)g(x) \geq f(x)$, hence the result. – Siong Thye Goh Dec 12 '17 at 23:47
• This is confusing. How can we separately calculate limits of f(x) & g(x) when limit of f(x) doesn't exist (i.e. it is infinite). ? – Anuj Dec 12 '17 at 23:48

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