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I am trying to calculate the following limit without L'Hôpital's rule:

$$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$

I tried every trick I know but nothing works. You don't have to prove it by definition.

Michael
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2 Answers2

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Let $\ln(n) = t$. We then have $n = e^t$. Hence, we have $$\lim_{n \to \infty} \dfrac{n}{\ln(n)} = \lim_{t \to \infty} \dfrac{e^t}{t}$$ Recall that $e^t = 1 + t + \dfrac{t^2}{2!} + \cdots > \dfrac{t^2}{2}$. Hence, we have $$\lim_{t \to \infty} \dfrac{e^t}{t} > \lim_{t \to \infty} \dfrac{t^2}{2t} = \lim_{t \to \infty} \dfrac{t}{2} = \infty$$

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Every time you make $n$ twice and big, you increase $\ln n$ by less than $1$ (since $e$ is less than $2$). So imagine repeatedly doubling the numerator and each time you do it, adding just $1$ to the denominator, and think about what that will approach. That does it.

PS inspired by a comment: $$ \frac{n}{\ln n} > \frac{n}{\log_2 n}. $$ A comparison test then says that if the latter goes to $\infty$, then so does the former.

Each value of $n$ is located between two powers of $2$: we have $2^m\le n\le 2^{m+1}$. That number $m$ is $m=\lfloor \log_2 n\rfloor$. So $m\le\log_2 n<m+1$. $$ \frac{2^m}{m+1}\le\frac{n}{m+1}<\frac{n}{\log_2 n}\le\frac n m. $$ So it is enough to show that $\dfrac{2^m}{m+1}\to\infty$. Show by induction that for some constant $c$, we have $$ \frac{2^m}{m+1} \ge \left(\frac 3 2\right)^m $$ and then do another comparison.

Michael Hardy
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