To evaluate the limit [tex]\(\lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 - 1}}\)[/tex], we need to use a method to handle the form involving infinity.
Here's a step-by-step approach:
1. Expression Analysis: We start with the expression [tex]\(\frac{2x - 1}{\sqrt{x^2 - 1}}\)[/tex].
2. Divide by x: To simplify the expression, we divide the numerator and the denominator by [tex]\(x\)[/tex]. This often helps in handling infinity limits because it normalizes the terms.
[tex]\[
\frac{2x - 1}{\sqrt{x^2 - 1}} = \frac{2x - 1}{x\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
3. Simplify the denominator: Notice that [tex]\(\sqrt{x^2 - 1} = x\sqrt{1 - \frac{1}{x^2}}\)[/tex].
So,
[tex]\[
\frac{2x - 1}{x \sqrt{1 - \frac{1}{x^2}}} = \frac{2x - 1}{x} \cdot \frac{1}{\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
4. Simplify the numerator: We simplify [tex]\(\frac{2x - 1}{x} = 2 - \frac{1}{x}\)[/tex].
5. Rewrite the expression: Thus, the expression becomes
[tex]\[
(2 - \frac{1}{x}) \cdot \frac{1}{\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
6. Evaluate the limit: As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{x} \to 0\)[/tex].
Therefore, the expression simplifies to:
[tex]\[
2 \cdot \frac{1}{\sqrt{1 - 0}} = 2 \cdot 1 = 2
\][/tex]
Thus, the limit is
[tex]\[
\lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 - 1}} = 2.
\][/tex]
The correct answer is (C) 2.
