Math Snap

STEP 1

1. The limit involves expressions with square roots and polynomial terms.
2. As $n$ approaches infinity, the dominant terms in the square roots will determine the behavior of the expression.
3. Rationalizing the expression may help simplify the limit calculation.

STEP 2

1. Simplify the expression by rationalizing the difference of square roots.
2. Evaluate the limit of the simplified expression as $n$ approaches infinity.

STEP 3

To simplify the expression, multiply and divide by the conjugate of the square roots:
$$\lim _{n \rightarrow \infty} \left( \sqrt{3 n^{2}+n} - \sqrt{3 n^{2}-5 n} \right)$$ Multiply and divide by the conjugate:
$$\lim _{n \rightarrow \infty} \frac{\left( \sqrt{3 n^{2}+n} - \sqrt{3 n^{2}-5 n} \right) \left( \sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n} \right)}{\sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n}}$$

STEP 4

Simplify the numerator using the difference of squares formula:
$$\left( \sqrt{3 n^{2}+n} \right)^2 - \left( \sqrt{3 n^{2}-5 n} \right)^2 = (3 n^{2} + n) - (3 n^{2} - 5 n)$$ Simplify further:
$$= 3n^{2} + n - 3n^{2} + 5n = 6n$$ The expression becomes:
$$\lim _{n \rightarrow \infty} \frac{6n}{\sqrt{3 n^{2}+n} + \sqrt{3 n^{2}-5 n}}$$

STEP 5

Factor out $n$ from the square roots in the denominator:
$$\sqrt{3 n^{2}+n} = n\sqrt{3 + \frac{1}{n}}$$ $$\sqrt{3 n^{2}-5 n} = n\sqrt{3 - \frac{5}{n}}$$ The expression becomes:
$$\lim _{n \rightarrow \infty} \frac{6n}{n\left(\sqrt{3 + \frac{1}{n}} + \sqrt{3 - \frac{5}{n}}\right)}$$ Cancel $n$ in the numerator and denominator:
$$\lim _{n \rightarrow \infty} \frac{6}{\sqrt{3 + \frac{1}{n}} + \sqrt{3 - \frac{5}{n}}}$$

Evaluate the limit as $n \rightarrow \infty$:
As $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$ and $\frac{5}{n} \rightarrow 0$, so:
$$\sqrt{3 + \frac{1}{n}} \rightarrow \sqrt{3}$$ $$\sqrt{3 - \frac{5}{n}} \rightarrow \sqrt{3}$$ Thus, the expression simplifies to:
$$\lim _{n \rightarrow \infty} \frac{6}{\sqrt{3} + \sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$$ The value of the limit is:
$$\boxed{\sqrt{3}}$$