<s>[INST] Sandwich Theorem not working? 
This is the limit I need to solve:
  $$\lim_{n \to \infty} \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4}$$

I simplified it to this:
$$\lim_{n \to \infty} \frac{2(4 \cos(n) - 3n^2)}{(6n^3 + 5n \sin(n))}.$$
At this point I want to use the Sandwich Theorem on the Numerator and Denominator to evaluate the limit. 
I use the fact that $\lim_{n \to \infty} \frac{a}{b} = \frac{\lim_{n \to \infty} a}{\lim_{n \to \infty} b}$ when  $b\ne 0$. 
By the Sandwich Theorem both the Numerator and Denominator is $\infty$.
Hence the answer is 1. 
But if I calculate the limit whole without splitting it into two I get $\frac{3}{2}$. Which answer is correct? Please Help!
 [/INST] We have that
$$\frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4  - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$\frac{(\pm4  - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\sim \frac{-6n^7}{6n^7} = -1$$
as already noticed by RobertZ, in a simpler way, we can directly use the same argument for the original limit.
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