Let's evaluate the limit [tex]\(\lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 - 1}}\)[/tex].
First, we can observe that as [tex]\( x \)[/tex] approaches infinity, the terms [tex]\(2x\)[/tex] and [tex]\(\sqrt{x^2}\)[/tex] will dominate in the numerator and the denominator, respectively. To make the expression easier to handle, let's analyze the behavior of both the numerator and the denominator separately.
The given expression is:
[tex]\[
\frac{2x - 1}{\sqrt{x^2 - 1}}
\][/tex]
To simplify this, let's divide the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\frac{\frac{2x - 1}{x}}{\frac{\sqrt{x^2 - 1}}{x}}
\][/tex]
This can be rewritten as:
[tex]\[
\frac{\frac{2x}{x} - \frac{1}{x}}{\frac{\sqrt{x^2 - 1}}{x}}
\][/tex]
Simplify the terms further:
[tex]\[
\frac{2 - \frac{1}{x}}{\sqrt{\frac{x^2 - 1}{x^2}}}
\][/tex]
Remember that [tex]\(\frac{x^2 - 1}{x^2}\)[/tex] can be expressed as:
[tex]\[
\frac{x^2 - 1}{x^2} = \frac{x^2}{x^2} - \frac{1}{x^2} = 1 - \frac{1}{x^2}
\][/tex]
So, our limit now becomes:
[tex]\[
\frac{2 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x^2}}}
\][/tex]
As [tex]\( x \)[/tex] approaches infinity, [tex]\(\frac{1}{x}\)[/tex] and [tex]\(\frac{1}{x^2}\)[/tex] approach 0. Thus, the expression simplifies to:
[tex]\[
\frac{2 - 0}{\sqrt{1 - 0}} = \frac{2}{1} = 2
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \to \infty} \frac{2x - 1}{\sqrt{x^2 - 1}} = 2
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{2}
\][/tex]
