Jtdylktuy

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 $bne 0$.

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!

You should revise your work. My advice is to apply the Sandwich Theorem in a different way.

Note that the given limit can be written as
$$lim_{n to infty} frac{n^2cdot (frac{4 cos(n)}{n^2} - 3)cdot n^5cdot(2 - frac{1}{n^2} + frac{1}{n^5})}{n^3cdot (6 + frac{5sin(n)}{n^2})cdot n^4cdot (1 + frac{2}{n})^4}$$
Simplify the powers of $n$ and recall that, just by the Sandwich Theorem, if $a_nto 0$ and $b_n$ is bounded then $lim_{nto infty}(a_ncdot b_n)=0$.

What is the final answer?

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.