I came across this question in an entrance exam of a local college where we are asked to evaluate the limit : $$\lim\limits_{n\to\infty} \frac{N_n}{\ln(n)}$$ Where $N_n$ denotes the number of digits of $n$, with the latter being a non zero positive integer. It seems I am lacking some sort of relationship between $N_n$ and $n$ (i.e equality/inequality). I could spot that $n \ge N_n$ but it doesn't seem useful in this case. Appreciate any help!
Asked
Active
Viewed 64 times
1
-
7Hint: try to relate $N_n$ to $\log_{10}n$. – lulu Jul 27 '20 at 22:15
-
@lulu can I just say, your comment game is on point – Integrand Jul 27 '20 at 22:26
-
@Integrand It's odd to think about which skills are enhanced by extended lockdown, and which deteriorate. Best to focus on the former, I find. – lulu Jul 27 '20 at 22:37
1 Answers
3
HINT
Following the suggestion given in the comments, we have that
$$\frac{\log_{10}n}{\ln(n)}\le \frac{N_n}{\ln(n)}\le \frac{1+\log_{10}n}{\ln(n)}$$
then we can conclude by squeeze theorem.
user
- 154,566
-
Thanks for the answer! one question: to show that inequality should I start with the expression of $n$ in base $10$ ? like $208=2.10^2+0.10^1+8.10^0$ ? or is there a simpler way ? – OUCHNA Jul 27 '20 at 22:33
-
-
$100\le 208<1000$ so between $10^2$ and $10^3$. Then notice all numbers $100..999<10^3$ all have $3$ digits. – zwim Jul 27 '20 at 22:46