Math Problem Statement

Évalue cette limite

Évalue cette limite: lim (x->∞) (sqrt(x+1) - sqrt(2x))

Solution

We are tasked with evaluating the limit:

limx(x+12x)\lim_{x \to \infty} \left( \sqrt{x + 1} - \sqrt{2x} \right)

To simplify this expression and find the limit, we can multiply by the conjugate of the expression. This will help eliminate the square roots:

limx(x+12x)x+1+2xx+1+2x\lim_{x \to \infty} \left( \sqrt{x + 1} - \sqrt{2x} \right) \cdot \frac{\sqrt{x + 1} + \sqrt{2x}}{\sqrt{x + 1} + \sqrt{2x}}

Multiplying both the numerator and denominator by the conjugate, we get:

=limx(x+1)2(2x)2x+1+2x= \lim_{x \to \infty} \frac{(\sqrt{x+1})^2 - (\sqrt{2x})^2}{\sqrt{x+1} + \sqrt{2x}}

Simplifying the terms in the numerator:

=limx(x+1)(2x)x+1+2x= \lim_{x \to \infty} \frac{(x+1) - (2x)}{\sqrt{x+1} + \sqrt{2x}} =limxx+12xx+1+2x= \lim_{x \to \infty} \frac{x + 1 - 2x}{\sqrt{x+1} + \sqrt{2x}} =limxx+1x+1+2x= \lim_{x \to \infty} \frac{-x + 1}{\sqrt{x+1} + \sqrt{2x}}

Now let's analyze this expression as xx \to \infty:

  • The numerator approaches x-x, since x+1x-x + 1 \approx -x for large xx.
  • The denominator approaches 2x\sqrt{2x}, because x+1x\sqrt{x + 1} \approx \sqrt{x} and x+1+2x2x+x\sqrt{x + 1} + \sqrt{2x} \approx \sqrt{2x} + \sqrt{x}.

Thus, the expression becomes:

limxx2x+x\lim_{x \to \infty} \frac{-x}{\sqrt{2x} + \sqrt{x}}

Factor x\sqrt{x} out of the denominator:

=limxxx(2+1)= \lim_{x \to \infty} \frac{-x}{\sqrt{x} (\sqrt{2} + 1)} =limxx2+1= \lim_{x \to \infty} \frac{-\sqrt{x}}{\sqrt{2} + 1}

As xx \to \infty, x\sqrt{x} \to \infty, so the entire expression tends to -\infty. Therefore, the limit is:

\boxed{-\infty}

Let me know if you want further details or have any questions.

Here are 5 questions to deepen your understanding:

  1. Why do we use the conjugate method when dealing with square roots in limits?
  2. How does factoring out x\sqrt{x} simplify the expression?
  3. What happens to the terms inside the square root as xx \to \infty?
  4. What are some other methods to evaluate limits with square roots?
  5. Can the limit of a square root expression approach a finite number when xx \to \infty?

Tip: Multiplying by the conjugate is a powerful tool for simplifying expressions involving square roots, especially when approaching limits at infinity.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Square Roots

Formulas

sqrt(x + 1) - sqrt(2x)

Theorems

Limit properties

Suitable Grade Level

University Level