To find the limit [tex]\(\lim_{{x \to \infty}} \frac{\sqrt{x^2 + 1}}{2x + 1}\)[/tex], let's analyze the expression step-by-step.
1. Simplify the Expression: Note that as [tex]\(x \to \infty\)[/tex], the term [tex]\(x^2\)[/tex] in the numerator will dominate over the constant [tex]\(1\)[/tex]. Thus, we need to factor out the dominant term [tex]\(x\)[/tex] inside the square root to simplify.
2. Rewrite the Numerator: Rewrite [tex]\(\sqrt{x^2 + 1}\)[/tex] by factoring [tex]\(x^2\)[/tex] out of the square root:
[tex]\[
\sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = x \sqrt{1 + \frac{1}{x^2}}
\][/tex]
3. Substitute the Simplified Form: Substitute the simplified form of the numerator back into the original expression:
[tex]\[
\frac{\sqrt{x^2 + 1}}{2x + 1} = \frac{x \sqrt{1 + \frac{1}{x^2}}}{2x + 1}
\][/tex]
4. Simplify the Rational Expression: To further simplify, divide both the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\frac{x \sqrt{1 + \frac{1}{x^2}}}{2x + 1} = \frac{x \sqrt{1 + \frac{1}{x^2}}}{x(2 + \frac{1}{x})} = \frac{\sqrt{1 + \frac{1}{x^2}}}{2 + \frac{1}{x}}
\][/tex]
5. Analyze the Simplified Expression as [tex]\(x \to \infty\)[/tex]:
- As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{x^2} \to 0\)[/tex]. Thus, [tex]\(\sqrt{1 + \frac{1}{x^2}} \to \sqrt{1} = 1\)[/tex].
- Similarly, as [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{x} \to 0\)[/tex]. Thus, [tex]\(2 + \frac{1}{x} \to 2\)[/tex].
6. Evaluate the Limit: Substituting these results into the simplified expression:
[tex]\[
\frac{\sqrt{1 + \frac{1}{x^2}}}{2 + \frac{1}{x}} \to \frac{1}{2}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{{x \to \infty}} \frac{\sqrt{x^2 + 1}}{2x + 1} = \frac{1}{2}
\][/tex]
