Math Snap

STEP 1

1. The expression involves two square roots and a limit as $n$ approaches infinity.
2. Simplification of the expression is necessary to evaluate the limit.
3. Rationalization techniques may be useful in simplifying the expression.

STEP 2

1. Simplify the expression inside the limit.
2. Rationalize the expression.
3. Evaluate the limit as $n$ approaches infinity.

STEP 3

First, observe the expression inside the limit:
$$\lim _{n \rightarrow \infty} \left( \sqrt{n^{4}+n^{2}} - \sqrt{n^{4}-5n^{2}+n} \right)$$ Notice that both terms under the square roots have $n^4$ as the leading term. We can factor $n^4$ out of each square root:
$$\sqrt{n^4(1 + \frac{1}{n^2})} - \sqrt{n^4(1 - \frac{5}{n^2} + \frac{1}{n^3})}$$ This simplifies to:
$$n^2 \sqrt{1 + \frac{1}{n^2}} - n^2 \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}$$

STEP 4

To simplify further, we rationalize the expression by multiplying and dividing by the conjugate:
$$\left( n^2 \sqrt{1 + \frac{1}{n^2}} - n^2 \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \right) \times \frac{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}$$ This results in:
$$\frac{n^2 \left( \left(1 + \frac{1}{n^2}\right) - \left(1 - \frac{5}{n^2} + \frac{1}{n^3}\right) \right)}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}$$ Simplifying the numerator:
$$n^2 \left( \frac{1}{n^2} + \frac{5}{n^2} - \frac{1}{n^3} \right) = n^2 \left( \frac{6}{n^2} - \frac{1}{n^3} \right) = 6 - \frac{1}{n}$$

Now, evaluate the limit as $n$ approaches infinity:
$$\lim _{n \rightarrow \infty} \frac{6 - \frac{1}{n}}{\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}}}$$ As $n \to \infty$, the terms $\frac{1}{n^2}$, $\frac{5}{n^2}$, and $\frac{1}{n^3}$ approach zero, so:
$$\sqrt{1 + \frac{1}{n^2}} \to 1$$ $$\sqrt{1 - \frac{5}{n^2} + \frac{1}{n^3}} \to 1$$ Thus, the denominator approaches $1 + 1 = 2$.
The limit becomes:
$$\lim _{n \rightarrow \infty} \frac{6 - \frac{1}{n}}{2} = \frac{6}{2} = 3$$ The value of the limit is:
$$\boxed{3}$$