Answer by NetUser5y62 for Limit of a sequence with floor function.
The answer to this question was sufficient for me to solve a similar problem. Just posting it here so others in future will find it useful.Show that for any $a\in \mathbb{R}$, the sequence of floor...
View ArticleAnswer by Eric for Limit of a sequence with floor function.
You can bound the floor function above and below by $n-1 \leq\lfloor n \rfloor \leq n$. This means you can bound your limit by$$\lim \frac{n+(n^\frac{1}{3}-1)^3}{n-\sqrt{n+9}} \leq \lim \frac{n+\lfloor...
View ArticleLimit of a sequence with floor function.
How do I compute the following limit: $\lim \limits_{n \to \infty} \frac{n +\lfloor \sqrt[3]n\rfloor^3}{n - \lfloor \sqrt{n+9}\rfloor}$Without the floor function this would be simple, but I never...
View Article
More Pages to Explore .....
