Calculate limit of Floor $\Rightarrow (n^2+n+1)*[\frac {2019}{n^2+n+1}]$ I can't figure out how to solve it.
The sequence is :
$(n^2+n+1)*[\frac {2019}{n^2+n+1}]$
I tried using the formula $x-1\lt [x] \le x$ and find the limit using the squeeze theorem.
so I get $\frac {2019}{n^2+n+1} -1 \lt [\frac {2019}{n^2+n+1}] \le \frac {2019}{n^2+n+1}$ , multyipling each side by ($n^2+n+1$).
$2019-(n^2+n+1) \lt (n^2+n+1)[\frac {2019}{n^2+n+1}] \le 2019$.
well , $\lim_{n\to \infty} (2019-(n^2+n+1)) = -\infty $.
$\lim_{n\to \infty} 2019 = 2019$
No common limit between them , so I can't use the squeeze theorem.
Thank you
You say ceiling, but you use the definition of floor. Ok, so:
if you meant floor
The floor of $[\frac{2019}{n^2+n+1}]=0$ is identically zero for $n$ big enough (say $n&gt;2019$). So your sequence looks like $a_1, \dots, a_{2019}, 0, 0, 0, 0, 0, \dots$
In particular, then, its limit for $n \to \infty$ is $0$.
if you meant ceiling
Here things are more interesting. Note that $n \leq [n]$. Now you have an indeterminate form because $[\frac{2019}{n^2+n+1}] =1$ but $[\lim_{n \to \infty} \frac{2019}{n^2+n+1}] = [0] = 0$.
What you can do is consider the following:
$$ (n^2+n+1) \frac{2019}{n^2+n+1} \leq (n^2+n+1) \left[\frac{2019}{n^2+n+1}\right]$$
which tells you that the limit cannot be zero, as the sequence is uniformly bigger than $2019$.
Now note that for big enough $n$ (just as before, take $n&gt;2019$),
$$(n^2+n+1) \left[\frac{2019}{n^2+n+1}\right] = n^2+n+1$$
since the quantity inside the brackets is nonzero, but certainly smaller than $1$, so its ceiling is $1$. In particular then your sequence gets arbitrarily big: for any number $M&gt;0$, there is $N$ such that for all $n&gt;N$ you have that $a_n &gt; M$, so the limit is $+\infty$.
Answer is Zero.
start with inside of brackets.as $n$ goes to infinity $\frac{2019}{n^2+n+1}$ goes to zero and is positive so its floor is absolute zero and we know that multiply of each number by absolute zero is zero.