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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)

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  • $\begingroup$ If you are ok, you can set as solved. Thanks! $\endgroup$
    – user
    Dec 15, 2017 at 21:42

3 Answers 3

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Simply note that:

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

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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.

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  • $\begingroup$ But that would require separating limits of numerator & denominator & considering them as 2 separate functions. Can we do that? $\endgroup$
    – Anuj
    Dec 12, 2017 at 23:35
  • $\begingroup$ check here point $4$. check the limit of the denominator is non-zero. $\endgroup$ Dec 12, 2017 at 23:36
  • $\begingroup$ 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 $\endgroup$
    – Anuj
    Dec 12, 2017 at 23:42
  • $\begingroup$ yes, when $x$ is huge, $f(x)g(x) \geq f(x)$, hence the result. $\endgroup$ Dec 12, 2017 at 23:47
  • $\begingroup$ 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). ? $\endgroup$
    – Anuj
    Dec 12, 2017 at 23:48
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Using equivalents for large $n$ (remembering that $\log(n)< n)$ $$\frac{1}{n-\log (n)}=\frac{1}{n}\frac{1}{1-\frac{\log (n)}{n}}\sim \frac{1}{n}\left(1+\frac{\log (n)}{n} \right)=\frac{1}{n}+\frac{\log (n)}{n^2}$$

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