Lim ( n → +∞ ) ( √(4n² + 3n + 1 - 2n) ) = a/b, tính √(a² + b²)

To solve the limit

\[
\lim_{n \to +\infty} \sqrt{4n^2 + 3n + 1 - 2n},
\]

we first simplify the expression inside the square root:

\[
4n^2 + 3n + 1 - 2n = 4n^2 + (3n - 2n) + 1 = 4n^2 + n + 1.
\]

Now we have:

\[
\lim_{n \to +\infty} \sqrt{4n^2 + n + 1}.
\]

Next, we can factor out \(n^2\) from the square root:

\[
\sqrt{4n^2 + n + 1} = \sqrt{n^2 \left(4 + \frac{n}{n^2} + \frac{1}{n^2}\right)} = n \sqrt{4 + \frac{1}{n} + \frac{1}{n^2}}.
\]

As \(n\) approaches infinity, \(\frac{1}{n}\) and \(\frac{1}{n^2}\) both approach 0:

\[
\lim_{n \to +\infty} n \sqrt{4 + \frac{1}{n} + \frac{1}{n^2}} = n \sqrt{4} = n \cdot 2 = 2n.
\]

Thus, we have:

\[
\lim_{n \to +\infty} \sqrt{4n^2 + 3n + 1 - 2n} = \lim_{n \to +\infty} 2n \text{ (since we only consider the leading term when } n \to \infty).
\]

As this approaches infinity, we can look back at the limit structure:

To find \(a\) and \(b\) in the form \( \frac{a}{b} \):

This expression is essentially \(2\), where \(a = 2\) and \(b = 1\).

Now, we calculate:

\[
\sqrt{a^2 + b^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}.
\]

Thus, the final answer is

\[
\sqrt{a^2 + b^2} = \sqrt{5}.
\]