Math Problem Statement

lim x -> infinity(sqrt(1 + x ^ 4) - 1 - x ^ 2)/(x ^ 2) =

A)-1

B0

C-2

D 2

Solution

We are tasked with evaluating the following limit:

limx1+x41x2x2\lim_{x \to \infty} \frac{\sqrt{1 + x^4} - 1 - x^2}{x^2}

Step 1: Simplify the Expression Inside the Limit

First, consider the term inside the square root:

1+x4\sqrt{1 + x^4}

For large values of xx, the term x4x^4 dominates over 11, so we can approximate:

1+x4x2asx\sqrt{1 + x^4} \approx x^2 \quad \text{as} \quad x \to \infty

Now rewrite the expression:

1+x41x2x2\frac{\sqrt{1 + x^4} - 1 - x^2}{x^2}

Step 2: Use Asymptotic Expansion

We can perform an asymptotic expansion for 1+x4\sqrt{1 + x^4} when xx is large. Let's expand 1+x4\sqrt{1 + x^4} as follows:

1+x4=x21+1x4x2(1+12x4)\sqrt{1 + x^4} = x^2 \sqrt{1 + \frac{1}{x^4}} \approx x^2 \left(1 + \frac{1}{2x^4}\right)

So, the approximation becomes:

1+x4x2+12x2\sqrt{1 + x^4} \approx x^2 + \frac{1}{2x^2}

Step 3: Substitute the Expansion into the Expression

Now substitute this approximation back into the original limit:

(x2+12x2)1x2x2\frac{\left(x^2 + \frac{1}{2x^2}\right) - 1 - x^2}{x^2}

Simplify the numerator:

(x2+12x21x2)=12x21\left(x^2 + \frac{1}{2x^2} - 1 - x^2\right) = \frac{1}{2x^2} - 1

So the expression becomes:

12x21x2\frac{\frac{1}{2x^2} - 1}{x^2}

Step 4: Simplify Further

Now break the expression into two terms:

12x41x2\frac{1}{2x^4} - \frac{1}{x^2}

Step 5: Take the Limit as xx \to \infty

As xx \to \infty, the term 12x4\frac{1}{2x^4} approaches 0, and 1x2\frac{1}{x^2} also approaches 0. Therefore, the limit simplifies to:

limx(12x41x2)=0\lim_{x \to \infty} \left( \frac{1}{2x^4} - \frac{1}{x^2} \right) = 0

Thus, the value of the limit is:

0\boxed{0}

The correct answer is B) 0.


Would you like more details or have any questions about the steps? Here are 5 related questions for further practice:

  1. How do you apply asymptotic expansions in limits?
  2. What other techniques can be used to evaluate limits involving square roots?
  3. How do you approach limits at infinity in general?
  4. Can you apply L'Hopital's Rule to this limit? What would the result be?
  5. How would you handle a similar limit with a different degree of polynomial inside the square root?

Tip: When working with limits involving square roots, try approximating the square root expression using a binomial expansion or leading-order terms to simplify the analysis.

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Math Problem Analysis

Mathematical Concepts

Limits
Asymptotic Expansion
Infinity
Algebraic Manipulation

Formulas

lim x -> infinity (sqrt(1 + x^4) - 1 - x^2)/(x^2)
Asymptotic expansion: sqrt(1 + x^4) ≈ x^2 + (1/2x^2)

Theorems

Limit Theorem
Asymptotic Approximation

Suitable Grade Level

College-Level Calculus