Evaluating the Limit of $\boldsymbol{(\sqrt{4n^2+n}-2n)}$ as $\boldsymbol{n \to \infty}$

The question asks us to find the value of the limit: $ L = \lim_{n\to\infty} (\sqrt{4n^2+n}-2n) $ This limit is an indeterminate form of the type $\infty - \infty$, because as $n$ approaches infinity, both $\sqrt{4n^2+n}$ and $2n$ grow infinitely large.

Handling Indeterminate Forms

To resolve this indeterminate form, we can use the technique of multiplying by the conjugate. The conjugate of $(\sqrt{4n^2+n}-2n)$ is $(\sqrt{4n^2+n}+2n)$. We multiply and divide the expression by its conjugate:

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{4n^2+n}$ and $b = 2n$, we get:

Simplifying the Expression

Now, the limit is in the form $\frac{\infty}{\infty}$, which is another indeterminate form. To evaluate this, we can divide both the numerator and the denominator by the highest power of $n$ in the denominator. The term $\sqrt{4n^2+n}$ behaves like $\sqrt{4n^2} = 2n$ for large $n$. So, the highest power is effectively $n$.

We divide the numerator by $n$ and the denominator by $n$. Remember that for $n > 0$, $n = \sqrt{n^2}$.

Final Limit Evaluation

Now we can evaluate the limit by substituting the limits of the individual terms. As $n \to \infty$, the term $\frac{1}{n}$ approaches 0.

Therefore, the value of the limit is $\frac{1}{4}$.