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?
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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 function $$ x_n = \frac{\lfloor a \times 10^{n-1}\rfloor}{10^{n-1}} $$ converges to $a$. Following the method above, we have $a-1\le \lfloor a\rfloor \le a$. Therefore
$$ \frac{a\times 10^{n-1}-1}{10^{n-1}}\le \frac{\lfloor a \times 10^{n-1}\rfloor}{10^{n-1}} \le \frac{a\times 10^{n-1}}{10^{n-1}} $$
By the Squeeze Theorem, the sequence $(x_n)\to a$.
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.