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 encountered it before so I have no idea what to do, maybe utilize the squeeze theorem somehow?
Best Answer
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
$$ \lim \frac{n+(n^\frac{1}{3}-1)^3}{n-\sqrt{n+9}} \leq \lim \frac{n+\lfloor n^\frac{1}{3} \rfloor^3}{n-\lfloor \sqrt{n+9}\rfloor } \leq \lim \frac{n+(n^\frac{1}{3})^3}{n-\sqrt{n+9}-1} $$ as in the left most limit you have made the numerator smaller and the denominator bigger, while in the right most limit you have done the opposite.
If you compute the limits of these two sequences (and show they are equal), then you get your limit.