To evaluate the limit [tex]\(\lim_{x \rightarrow \infty} \frac{2x - 1}{\sqrt{x^2 - 1}}\)[/tex], we can use algebraic manipulation to simplify the expression. Here's the step-by-step solution:
1. Rewrite the expression inside the limit:
[tex]\[
\lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 - 1}}
\][/tex]
2. Factor [tex]\(x^2\)[/tex] out of the square root in the denominator:
To do this, notice that:
[tex]\[
\sqrt{x^2 - 1} = \sqrt{x^2 \left( 1 - \frac{1}{x^2} \right)} = x \sqrt{1 - \frac{1}{x^2}}
\][/tex]
Thus, the expression becomes:
[tex]\[
\lim_{x \to \infty} \frac{2x - 1}{x \sqrt{1 - \frac{1}{x^2}}}
\][/tex]
3. Simplify the fraction:
Divide both the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\lim_{x \to \infty} \frac{2x - 1}{x \sqrt{1 - \frac{1}{x^2}}} = \lim_{x \to \infty} \frac{\frac{2x - 1}{x}}{\sqrt{1 - \frac{1}{x^2}}} = \lim_{x \to \infty} \frac{2 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
4. Evaluate the limit as [tex]\(x \to \infty\)[/tex]:
- As [tex]\(x \rightarrow \infty\)[/tex], [tex]\(\frac{1}{x} \rightarrow 0\)[/tex].
- Therefore, the numerator [tex]\(2 - \frac{1}{x} \rightarrow 2\)[/tex].
- The term inside the square root, [tex]\(1 - \frac{1}{x^2} \rightarrow 1\)[/tex].
- Thus, the square root [tex]\(\sqrt{1 - \frac{1}{x^2}} \rightarrow \sqrt{1} = 1\)[/tex].
Putting these results together:
[tex]\[
\lim_{x \to \infty} \frac{2 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x^2}}} = \frac{2 - 0}{1} = 2
\][/tex]
Therefore, the limit is [tex]\(\boxed{2}\)[/tex].
