We are tasked with computing the limit of the given expression as $n \to \infty$:

$\lim_{n \to \infty} \left( \sqrt{4 + n^2} - \sqrt{n^2 + 1} \right)$

To solve this, we can simplify the expression by using some algebraic manipulation. Specifically, we'll multiply and divide by the conjugate to eliminate the square roots.

Step 1: Multiply and divide by the conjugate

We multiply the given expression by the conjugate of the two square roots:

$\left( \sqrt{4 + n^2} - \sqrt{n^2 + 1} \right) \cdot \frac{\sqrt{4 + n^2} + \sqrt{n^2 + 1}}{\sqrt{4 + n^2} + \sqrt{n^2 + 1}}$

This simplifies the difference of square roots:

$\frac{\left( \sqrt{4 + n^2} - \sqrt{n^2 + 1} \right) \cdot \left( \sqrt{4 + n^2} + \sqrt{n^2 + 1} \right)}{\sqrt{4 + n^2} + \sqrt{n^2 + 1}} = \frac{(4 + n^2) - (n^2 + 1)}{\sqrt{4 + n^2} + \sqrt{n^2 + 1}}$

Step 2: Simplify the numerator

Now simplify the numerator:

$(4 + n^2) - (n^2 + 1) = 4 + n^2 - n^2 - 1 = 3$

So the expression becomes:

$\frac{3}{\sqrt{4 + n^2} + \sqrt{n^2 + 1}}$

Step 3: Evaluate the limit

As $n \to \infty$, both $\sqrt{4 + n^2}$ and $\sqrt{n^2 + 1}$ behave like $n$, since $n^2$ dominates both the terms inside the square roots. More precisely:

$\sqrt{4 + n^2} \sim n \quad \text{and} \quad \sqrt{n^2 + 1} \sim n$

Thus, for large $n$, the denominator becomes approximately:

$\sqrt{4 + n^2} + \sqrt{n^2 + 1} \sim n + n = 2n$

Now, the expression becomes:

$\frac{3}{2n}$

Step 4: Take the limit

As $n \to \infty$, $\frac{3}{2n} \to 0$.

Therefore, the limit of the given expression is:

$\boxed{0}$

Would you like further clarification or details on any step?

Here are 5 related questions:

1. How do you handle limits involving square roots and infinity?
2. Why is multiplying by the conjugate a useful technique in this problem?
3. Can you explain why $\sqrt{n^2 + a}$ behaves like $n$ as $n \to \infty$?
4. What happens if the term inside the square root grows faster than $n^2$?
5. How would the result change if the constants inside the square roots were different?

Tip:

When computing limits involving square roots, try factoring out the dominant term from the square root or using algebraic tricks like multiplying by the conjugate to simplify the expression.