To find the limits, we need to analyze each expression as \( n \to \infty \).

a) \(\lim_{n \to \infty} \left(\sqrt{4n^2 + 5n - 2n} \right)\)

Simplify the expression inside the square root:

\[
= \lim_{n \to \infty} \sqrt{4n^2 + 3n}
\]

Factor out \( n^2 \) from the square root:

\[
= \lim_{n \to \infty} \sqrt{n^2(4 + \frac{3}{n})} = \lim_{n \to \infty} n \sqrt{4 + \frac{3}{n}}
\]

As \( n \to \infty \), \(\frac{3}{n} \to 0\):

\[
= \lim_{n \to \infty} n \sqrt{4} = \lim_{n \to \infty} 2n = \infty
\]

b) \(\lim_{n \to \infty} \left(\sqrt{2n+1} - \sqrt{n} \right)\)

We rationalize by multiplying by the conjugate:

\[
= \lim_{n \to \infty} \frac{(\sqrt{2n+1} - \sqrt{n})(\sqrt{2n+1} + \sqrt{n})}{\sqrt{2n+1} + \sqrt{n}}
= \lim_{n \to \infty} \frac{(2n+1) - n}{\sqrt{2n+1} + \sqrt{n}}
\]

Simplify:

\[
= \lim_{n \to \infty} \frac{n+1}{\sqrt{2n+1} + \sqrt{n}}
\]

Divide the numerator and denominator by \(\sqrt{n}\):

\[
= \lim_{n \to \infty} \frac{\sqrt{n} + \frac{1}{\sqrt{n}}}{\sqrt{2 + \frac{1}{n}} + 1}
= \frac{1}{\sqrt{2} + 1}
\]

Hence, the limit is:

\[
\frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \sqrt{2} - 1
\]

c) \(\lim_{n \to \infty} \left(3n - \sqrt{9n^2 + 1} \right)\)

Factor:

\[
= \lim_{n \to \infty} \left(3n - \sqrt{n^2(9 + \frac{1}{n^2})} \right)
= \lim_{n \to \infty} \left(3n - n\sqrt{9 + \frac{1}{n^2}} \right)
\]

Simplify:

\[
= \lim_{n \to \infty} n \left(3 - \sqrt{9 + \frac{1}{n^2}} \right)
\]

Evaluate the limit:

\[
= \lim_{n \to \infty} n \cdot \frac{9 - (9 + \frac{1}{n^2})}{3 + \sqrt{9 + \frac{1}{n^2}}}
= \lim_{n \to \infty} \left(-\frac{1}{n(3 + \sqrt{9 + \frac{1}{n^2}})}\right) \to 0
\]

d) \(\lim_{n \to \infty} \left(\sqrt[3]{n^3 - 2n - n} \right)\)

Simplify:

\[
= \lim_{n \to \infty} \sqrt[3]{n^3 - 3n}
\]

Factor out \( n^3 \):

\[
= \lim_{n \to \infty} n \sqrt[3]{1 - \frac{3}{n^2}}
\]

As \( n \to \infty \), \(\frac{3}{n^2} \to 0\):

\[
= \lim_{n \to \infty} n \times 1 = \infty
\]

Summary of limits:
a) \(\infty\)
b) \(\sqrt{2} - 1\)
c) \(0\)
d) \(\infty\)