Math Problem Statement

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

Solution

We are tasked with finding the limit:

limn(n2+4nn)\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:

limn(n2+4nn)×n2+4n+nn2+4n+n\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:

=limn(n2+4nn)(n2+4n+n)n2+4n+n= \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:

(n2+4n)2n2=(n2+4n)n2=4n\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:

limn4nn2+4n+n\lim_{n \to \infty} \frac{\frac{4}{n}}{\sqrt{n^2 + \frac{4}{n}} + n}

Step 3: Simplify the denominator

As nn \to \infty, we can factor out nn from the square root in the denominator:

n2+4n=n1+4n3\sqrt{n^2 + \frac{4}{n}} = n \sqrt{1 + \frac{4}{n^3}}

So the expression becomes:

limn4nn(1+4n3+1)\lim_{n \to \infty} \frac{\frac{4}{n}}{n \left( \sqrt{1 + \frac{4}{n^3}} + 1 \right)}

Step 4: Evaluate the limit

As nn \to \infty, 4n30\frac{4}{n^3} \to 0, so 1+4n31\sqrt{1 + \frac{4}{n^3}} \to 1. Therefore, the expression simplifies to:

limn4nn(1+1)=limn4n2n=limn42n2=0\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:

0\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 limn(n2+knn)\lim_{n \to \infty} \left( \sqrt{n^2 + \frac{k}{n}} - n \right) for any constant kk?
  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.

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Math Problem Analysis

Mathematical Concepts

Limits
Conjugate Method
Simplification of Expressions at Infinity

Formulas

lim_{n \to \infty} \left( \sqrt{n^2 + \frac{4}{n}} - n \right)
Difference of Squares: (a - b)(a + b) = a^2 - b^2
Factoring Square Roots: \sqrt{n^2 + \frac{4}{n}} = n \sqrt{1 + \frac{4}{n^3}}

Theorems

Limit at Infinity
Asymptotic Behavior of Functions
Squeeze Theorem (implicitly used in understanding limit behavior)

Suitable Grade Level

Grades 11-12 (Advanced High School Calculus)