Question

The value of \( \lim_{n \to \infty} (\sqrt{4n^2+n} - 2n) \) is:

• A. \(\frac{1}{2}\)
• B. 0
• C. \(\frac{1}{4}\)
• D. 1

Hint:

When faced with a limit of the form \(\sqrt{an^2+bn+c} - pn\), multiplying by the conjugate is the go-to strategy. It reliably transforms the \(\infty-\infty\) form into a more manageable \(\frac{\infty}{\infty}\) form, which can then be solved by dividing by the highest power of \(n\).

Answer 1

The Correct Option is C

Step 1: Problem Overview:
The given limit has an indeterminate form of \(\infty - \infty\). To solve this, which involves square roots, we multiply by the conjugate.

Step 2: Method:
We use the identity \((a-b)(a+b) = a^2 - b^2\) and multiply by \(\frac{\sqrt{4n^2+n} + 2n}{\sqrt{4n^2+n} + 2n}\).

Step 3: Solution:
\[ \lim_{n \to \infty} (\sqrt{4n^2+n} - 2n) = \lim_{n \to \infty} (\sqrt{4n^2+n} - 2n) \times \frac{\sqrt{4n^2+n} + 2n}{\sqrt{4n^2+n} + 2n} \]\[ = \lim_{n \to \infty} \frac{(\sqrt{4n^2+n})^2 - (2n)^2}{\sqrt{4n^2+n} + 2n} = \lim_{n \to \infty} \frac{(4n^2+n) - 4n^2}{\sqrt{4n^2+n} + 2n} \]\[ = \lim_{n \to \infty} \frac{n}{\sqrt{4n^2+n} + 2n} \]The limit is now in the form \(\frac{\infty}{\infty}\). We divide the numerator and denominator by \(n\).\[ = \lim_{n \to \infty} \frac{n/n}{(\sqrt{4n^2+n})/n + (2n)/n} = \lim_{n \to \infty} \frac{1}{\sqrt{4n^2/n^2+n/n^2} + 2} \]\[ = \lim_{n \to \infty} \frac{1}{\sqrt{4 + 1/n} + 2} \]As \(n \to \infty\), \(1/n \to 0\).\[ = \frac{1}{\sqrt{4 + 0} + 2} = \frac{1}{2 + 2} = \frac{1}{4} \]
Step 4: Result:
The limit's value is \(\frac{1}{4}\).