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.

  • $\begingroup$ perhaps you mean $\lim _{ n\rightarrow \infty }\frac { nln(n) }{ nlog(n) } $ $\endgroup$ – haqnatural Apr 11 '17 at 18:13
  • 5
    $\begingroup$ L'Hôpital? Great Scot! Why not simply use that $$\frac{n\ln n}{n\log n}=\frac{\ln n}{\log n}$$ and remember that $$\log_{10}n=\frac{\ln n}{\ln10}$$ $\endgroup$ – Did Apr 11 '17 at 18:13
  • $\begingroup$ 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. $\endgroup$ – 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 :

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

  • $\begingroup$ Why a downvote? $\endgroup$ – 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!

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.