find the limit of 
lim n > inf (sqrt(n^2+4/n)-n)

We are tasked with finding the limit:

\[
\lim_{n \to \infty} \left( \sqrt{n^2 + \frac{4}{n}} - n \right)
\]

### Step 1: Rewrite the expression
We can manipulate the given expression to simplify it. Begin by multiplying and dividing by the conjugate:

\[
\lim_{n \to \infty} \left( \sqrt{n^2 + \frac{4}{n}} - n \right) \times \frac{\sqrt{n^2 + \frac{4}{n}} + n}{\sqrt{n^2 + \frac{4}{n}} + n}
\]

This gives:

\[
= \lim_{n \to \infty} \frac{\left( \sqrt{n^2 + \frac{4}{n}} - n \right) \left( \sqrt{n^2 + \frac{4}{n}} + n \right)}{\sqrt{n^2 + \frac{4}{n}} + n}
\]

### Step 2: Simplify the numerator
The numerator becomes a difference of squares:

\[
\left( \sqrt{n^2 + \frac{4}{n}} \right)^2 - n^2 = \left( n^2 + \frac{4}{n} \right) - n^2 = \frac{4}{n}
\]

So, the expression is now:

\[
\lim_{n \to \infty} \frac{\frac{4}{n}}{\sqrt{n^2 + \frac{4}{n}} + n}
\]

### Step 3: Simplify the denominator
As \(n \to \infty\), we can factor out \(n\) from the square root in the denominator:

\[
\sqrt{n^2 + \frac{4}{n}} = n \sqrt{1 + \frac{4}{n^3}}
\]

So the expression becomes:

\[
\lim_{n \to \infty} \frac{\frac{4}{n}}{n \left( \sqrt{1 + \frac{4}{n^3}} + 1 \right)}
\]

### Step 4: Evaluate the limit
As \(n \to \infty\), \(\frac{4}{n^3} \to 0\), so \(\sqrt{1 + \frac{4}{n^3}} \to 1\). Therefore, the expression simplifies to:

\[
\lim_{n \to \infty} \frac{\frac{4}{n}}{n (1 + 1)} = \lim_{n \to \infty} \frac{\frac{4}{n}}{2n} = \lim_{n \to \infty} \frac{4}{2n^2} = 0
\]

### Final Answer:
The limit is:

\[
\boxed{0}
\]

Would you like more details or have any other questions?

---

Here are five related questions to explore further:
1. What is the limit of \(\lim_{n \to \infty} \left( \sqrt{n^2 + \frac{k}{n}} - n \right)\) for any constant \(k\)?
2. How does rationalizing expressions help in finding limits?
3. What are the general techniques to evaluate limits at infinity for complex expressions?
4. Can the same limit technique be used for functions involving trigonometric expressions?
5. How do higher-order terms affect the behavior of limits?

**Tip**: When dealing with square roots in limits, try multiplying by the conjugate to simplify the expression.

find the limit of lim n > inf (sqrt(n^2+4/n)-n)

This detailed step-by-step guide helps you find the limit of lim n → ∞ (sqrt(n^2 + 4/n) - n). Learn about simplifying complex expressions using the conjugate method and factoring at infinity. Suitable for advanced high school calculus students.

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