# Limit of $n\ln(n) / n\log(n)$

I am beginning to learn about limits in my calculus class and I have been working on some of the practice problems. However, there is one that I am getting stuck on. The problem is:

$$\lim \limits_{x \to \infty} \frac{n\ln(n)}{n\log(n)}.$$

I'm assuming I need to use L'Hopital's rule, which would get rid of both of the $n$ terms, but I am not sure how to apply L'Hopital's rule to a logarithmic expression.

• perhaps you mean $\lim _{ n\rightarrow \infty }\frac { nln(n) }{ nlog(n) }$ – haqnatural Apr 11 '17 at 18:13
• L'Hôpital? Great Scot! Why not simply use that $$\frac{n\ln n}{n\log n}=\frac{\ln n}{\log n}$$ and remember that $$\log_{10}n=\frac{\ln n}{\ln10}$$ – Did Apr 11 '17 at 18:13
• Don't need lhopital to "cancel" ns. And $\log n = \frac {\ln n}{\ln 10}$ is a natural ratio to make this limit really easy. $\lim \frac {n\ln n}{n\log n}=\lim \frac {\ln n}{\log n} = \lim \frac {\ln n}{\frac {\ln n}{\ln 10}} = \lim \ln 10 = \ln 10$. But if you want l'hpital why would logarithmic expressions be any different $(n\ln n)' = \ln n + \frac nn$ and $(n\log n)' = \log n + n*\ln (10) \frac 1n$. So do it. – fleablood Apr 11 '17 at 18:19

Recall property of logarithms : $$\log_{n}m =\frac{\log_{a}m}{\log_{a}n}$$

Therefore; $\ln n = \log_e n = \dfrac{\log_{10} n}{\log_{10} e} \implies \dfrac{\ln n}{\log_{10} n}= \dfrac{1}{\log_{10}e} = \ln 10$

$$\dfrac{n \ln(n)}{n\log(n)} = \ln10$$

Since the function is constant Limit Doesn't matter,

Therefore :

$$\lim_{n \to \infty}\dfrac{n \ln(n)}{n\log(n)} = \ln10 \approx 2.303$$

• Why a downvote? – Jaideep Khare Apr 12 '17 at 13:22

1) Using L'hopital to get rid of $n$s doesn't work

$\lim \frac {nf(n)}{ng(n)} = \lim \frac {nf'(n) + f(n)}{ng'(n) + g(n)}$ doesn't make anything any easier.

2) You don't know how to use L'hopitals with log exapressions? Why not? They are no different.

$\frac {d\ln x}{d x }= \frac 1x$ and as $\log_{10} x = \log_{10} e^{\ln x} = \ln x*\log_{10} e$ we have $\frac {d\log x}{dx} = \frac 1x\log e$.

So by L'hoptital

$\lim \frac {n\ln n}{n \log n} = \lim \frac{\ln n + \frac nn}{\log n + \log e \frac nn} = \lim \frac {\ln n + 1}{\log n + 1}$.

Which doesn't help us in the least.

[But remember $\log_b a = \frac {\log_c a}{\log_c b}$ so $\log n = \frac {\ln n}{\ln 10}$ and $\ln n = \frac {\log n}{\log e}$ and so $\log e = \frac 1{\ln 10}$.]

3) You cancel $n$ not by L'Hoptital but... by canceling $n$s!

$\lim_{n\rightarrow \infty} \frac {n\ln n}{n \log n} = \lim_{n \ne 0;n\rightarrow \infty} \frac {n\ln n}{n \log n} = \lim_{n\rightarrow \infty} \frac {\ln n}{\log n} = \lim_{n\ne 1; n\rightarrow \infty} \frac {\ln n}{\frac {\ln n}{\ln 10}} = \lim \frac {1}{\frac {1}{\ln 10}} = \ln 10$.