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Пользуясь арифметическими свойствами, вычислите


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.

Пользуясь арифметическими свойствами, вычислите: lim (n → ∞) (√(4 + n²) - √(n² + 1))

Explore the detailed step-by-step solution for calculating the limit lim (n → ∞) (√(4 + n²) - √(n² + 1)). Learn about algebraic manipulation techniques, including multiplying by the conjugate and analyzing the behavior of square roots at large values of n.

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