Question

The value of \[ \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right) \] is equal to:

• A. -1
• B. -0.5
• C. -2
• D. 0

Hint:

When dealing with limits involving square roots, multiplying by the conjugate can often help simplify the expression and make it easier to evaluate.

Answer 1

The Correct Option is B

We are asked to find the limit of the expression as \( x \to \infty \). First, let's simplify the expression: \[ \lim_{x \to \infty} \left( x - \sqrt{x^2 + x} \right). \] To simplify, multiply and divide by the conjugate: \[ x - \sqrt{x^2 + x} = \frac{\left( x - \sqrt{x^2 + x} \right)\left( x + \sqrt{x^2 + x} \right)}{x + \sqrt{x^2 + x}}. \] This simplifies to: \[ \frac{x^2 - (x^2 + x)}{x + \sqrt{x^2 + x}} = \frac{-x}{x + \sqrt{x^2 + x}}. \] Now, factor \( x \) from both the numerator and denominator: \[ \frac{-x}{x \left( 1 + \sqrt{1 + \frac{1}{x}} \right)} = \frac{-1}{1 + \sqrt{1 + \frac{1}{x}}}. \] As \( x \to \infty \), \( \frac{1}{x} \to 0 \), so the expression simplifies to: \[ \frac{-1}{1 + \sqrt{1}} = \frac{-1}{2}. \] Thus, the value of the limit is \( -0.5 \), which corresponds to option (B).