Math Problem Statement

Evaluate \( \lim_{{x \to \infty}} \frac{\sqrt{x+1} - \sqrt{x}}{1} \approx \frac{1}{2} \)

Solution

The expression you've provided is:

limxx+1x1\lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1}

and it seems to imply that the limit is equal to 12\frac{1}{2}.

Let's solve this step-by-step.

Step-by-Step Solution

  1. Original Expression: limxx+1x1\lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1}

  2. Multiply by the Conjugate: To simplify, we multiply the numerator and the denominator by the conjugate of the numerator, which is x+1+x\sqrt{x + 1} + \sqrt{x}: limx(x+1x)(x+1+x)1(x+1+x)\lim_{{x \to \infty}} \frac{\left(\sqrt{x + 1} - \sqrt{x}\right) \cdot \left(\sqrt{x + 1} + \sqrt{x}\right)}{1 \cdot \left(\sqrt{x + 1} + \sqrt{x}\right)}

  3. Simplify the Numerator: Using the difference of squares, (ab)(a+b)=a2b2(a - b)(a + b) = a^2 - b^2, we get: =limx(x+1)xx+1+x= \lim_{{x \to \infty}} \frac{(x + 1) - x}{\sqrt{x + 1} + \sqrt{x}}

    Simplifying further, the xx terms cancel: =limx1x+1+x= \lim_{{x \to \infty}} \frac{1}{\sqrt{x + 1} + \sqrt{x}}

  4. Evaluate the Denominator: As xx approaches infinity, both x+1\sqrt{x + 1} and x\sqrt{x} behave like x\sqrt{x}. So: x+1+x2x\sqrt{x + 1} + \sqrt{x} \approx 2\sqrt{x}

    Thus, the expression becomes: limx12x\lim_{{x \to \infty}} \frac{1}{2\sqrt{x}}

  5. Take the Limit: As xx \to \infty, 12x0\frac{1}{2\sqrt{x}} \to 0.

Conclusion

The limit is: limxx+1x1=0\lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1} = 0

Therefore, the correct answer is 00, not 12\frac{1}{2}.

Would you like more details or have any questions?


Here are five related questions to deepen understanding:

  1. Why do we use the conjugate method when simplifying expressions with square roots in limits?
  2. How does the behavior of x\sqrt{x} as xx approaches infinity impact limits?
  3. What other techniques, besides conjugate multiplication, can be used for evaluating limits involving radicals?
  4. How would the limit change if it was x+cx\sqrt{x + c} - \sqrt{x} for some constant cc?
  5. What is the significance of higher-order terms when evaluating limits of functions at infinity?

Tip: When dealing with limits at infinity, always consider simplifying expressions using techniques like conjugate multiplication or factoring dominant terms.

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Math Problem Analysis

Mathematical Concepts

Limits
Conjugate multiplication
Behavior of functions at infinity

Formulas

\( \lim_{{x \to \infty}} \frac{\sqrt{x + 1} - \sqrt{x}}{1} \)
Difference of squares: \( (a - b)(a + b) = a^2 - b^2 \)

Theorems

Limit theorem for infinity
Difference of squares

Suitable Grade Level

Grades 11-12