As x approaches infinity, what is the limit of the square root of (x^2 - 3)/(2x + 1)?

As x approaches infinity, what is the limit of the square root of (x^2 - 3)/(2x + 1)?

• A. 0
• B. 1
• C. 1/2
• D. Infinity

Answer: (C)

To find the limit of the expression \(\sqrt{\frac{x^2 - 3}{2x + 1}}\) as \(x\) approaches infinity, we can simplify the determination of the limit by dividing both the numerator and the denominator by \(x\), the highest power in the denominator. Starting with the expression: \[ \sqrt{\frac{x^2 - 3}{2x + 1}} = \sqrt{\frac{x^2(1 - \frac{3}{x^2})}{x(2 + \frac{1}{x})}} \] As \(x\) approaches infinity, \(\frac{3}{x^2}\) approaches 0 and \(\frac{1}{x}\) also approaches 0. This leads us to simplify the expression further: \[ = \sqrt{\frac{x^2(1 - 0)}{x(2 + 0)}} = \sqrt{\frac{x^2}{2x}} = \sqrt{\frac{x}{2}} = \frac{\sqrt{x}}{\sqrt{2}} \] As \(x\) approaches infinity, the term \(\sqrt{x}\) increases without bound. The limit can

To find the limit of the expression (\sqrt{\frac{x^2 - 3}{2x + 1}}) as (x) approaches infinity, we can simplify the determination of the limit by dividing both the numerator and the denominator by (x), the highest power in the denominator.

Starting with the expression:

[

\sqrt{\frac{x^2 - 3}{2x + 1}} = \sqrt{\frac{x^2(1 - \frac{3}{x^2})}{x(2 + \frac{1}{x})}}

]

As (x) approaches infinity, (\frac{3}{x^2}) approaches 0 and (\frac{1}{x}) also approaches 0. This leads us to simplify the expression further:

[

= \sqrt{\frac{x^2(1 - 0)}{x(2 + 0)}} = \sqrt{\frac{x^2}{2x}} = \sqrt{\frac{x}{2}} = \frac{\sqrt{x}}{\sqrt{2}}

]

As (x) approaches infinity, the term (\sqrt{x}) increases without bound. The limit can