Math Problem Statement

Find the limit: lim(x -> +∞) of the expression (x - sqrt(x^2 + x + 1)) / (2x + sqrt(4x^2 + x))

Solution

The problem in the image is the following limit expression:

limx+xx2+x+12x+4x2+x\lim_{{x \to +\infty}} \frac{x - \sqrt{x^2 + x + 1}}{2x + \sqrt{4x^2 + x}}

To solve this limit as x+x \to +\infty, we use asymptotic analysis and simplifying techniques. The key idea is to factor out terms that dominate as xx \to \infty and deal with the square roots.

Step-by-step solution:

1. Focus on the numerator:

xx2+x+1x - \sqrt{x^2 + x + 1}

Factor out xx from the square root: x2+x+1=x1+1x+1x2\sqrt{x^2 + x + 1} = x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} As xx \to \infty, the terms 1x\frac{1}{x} and 1x2\frac{1}{x^2} become negligible, so: x2+x+1x(1+12x)\sqrt{x^2 + x + 1} \sim x \left( 1 + \frac{1}{2x} \right) Thus, the expression becomes: xx(1+12x)=xx12=12x - x \left( 1 + \frac{1}{2x} \right) = x - x - \frac{1}{2} = -\frac{1}{2} So, the numerator approaches 12-\frac{1}{2} as xx \to \infty.

2. Focus on the denominator:

2x+4x2+x2x + \sqrt{4x^2 + x} Again, factor out xx from the square root: 4x2+x=2x1+14x\sqrt{4x^2 + x} = 2x \sqrt{1 + \frac{1}{4x}} As xx \to \infty, 14x\frac{1}{4x} becomes negligible, so: 4x2+x2x(1+18x)\sqrt{4x^2 + x} \sim 2x \left( 1 + \frac{1}{8x} \right) Thus, the denominator becomes: 2x+2x(1+18x)=2x+2x+14=4x+142x + 2x \left( 1 + \frac{1}{8x} \right) = 2x + 2x + \frac{1}{4} = 4x + \frac{1}{4} So, the denominator approaches 4x4x as xx \to \infty.

3. Combine the results:

Now, the limit can be simplified: limx+124x=0\lim_{{x \to +\infty}} \frac{-\frac{1}{2}}{4x} = 0

Final answer:

limx+xx2+x+12x+4x2+x=0\lim_{{x \to +\infty}} \frac{x - \sqrt{x^2 + x + 1}}{2x + \sqrt{4x^2 + x}} = 0

Would you like more details on any part of the process?

Related questions:

  1. How does the technique of factoring out dominant terms help in limits involving square roots?
  2. What are the most common ways to simplify expressions involving large xx?
  3. Why do terms like 1x\frac{1}{x} and 1x2\frac{1}{x^2} become negligible as xx \to \infty?
  4. Can this limit be solved using L'Hôpital's Rule instead? How?
  5. What are the potential pitfalls of not simplifying correctly in limits with square roots?

Tip:

When dealing with limits as xx \to \infty, always consider the highest-order terms first, as lower-order terms often become negligible.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Asymptotic Analysis
Square Roots Simplification

Formulas

lim(x -> +∞) of f(x) / g(x) when dealing with dominant terms
sqrt(a^2 + b) ≈ a(1 + b/(2a^2)) for large a

Theorems

Simplification of limits by factoring dominant terms
Asymptotic approximations for square roots

Suitable Grade Level

Undergraduate/Graduate