To solve the limit [tex]\(\lim_{x \rightarrow \infty} \frac{2x - 1}{\sqrt{x^2 - 1}}\)[/tex], let's go through the solution step-by-step.
First, we observe that as [tex]\(x\)[/tex] approaches infinity, both the numerator [tex]\(2x - 1\)[/tex] and the denominator [tex]\(\sqrt{x^2 - 1}\)[/tex] grow without bound. Therefore, we need to simplify the expression so that we can better understand the behavior of the function as [tex]\(x\)[/tex] becomes very large.
Let's rewrite the given expression, focusing on simplifying it as [tex]\(x\)[/tex] becomes very large:
[tex]\[
\frac{2x - 1}{\sqrt{x^2 - 1}}
\][/tex]
Divide the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\frac{2x - 1}{\sqrt{x^2 - 1}} = \frac{x(2 - \frac{1}{x})}{\sqrt{x^2(1 - \frac{1}{x^2})}}
\][/tex]
Simplify the denominator:
[tex]\[
\sqrt{x^2(1 - \frac{1}{x^2})} = x \sqrt{1 - \frac{1}{x^2}}
\][/tex]
Now the expression becomes:
[tex]\[
\frac{x(2 - \frac{1}{x})}{x \sqrt{1 - \frac{1}{x^2}}}
\][/tex]
Since [tex]\(x \neq 0\)[/tex], we can cancel out [tex]\(x\)[/tex] from the numerator and the denominator:
[tex]\[
\frac{2 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
Now, evaluate the limit as [tex]\(x\)[/tex] approaches infinity. As [tex]\(x\)[/tex] becomes very large, the terms [tex]\(\frac{1}{x}\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex] approach 0:
[tex]\[
\lim_{x \rightarrow \infty} \frac{2 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x^2}}} = \frac{2 - 0}{\sqrt{1 - 0}} = \frac{2}{1} = 2
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{2}
\][/tex]
