To find the limit of the function [tex]\(\frac{\sqrt{2x + 3}}{x + 3}\)[/tex] as [tex]\(x\)[/tex] approaches 3, we follow these steps:
1. Substitution: First, we substitute [tex]\(x = 3\)[/tex] directly into the function to see if it gives us a clear value or an indeterminate form.
[tex]\[
\frac{\sqrt{2(3) + 3}}{3 + 3} = \frac{\sqrt{6 + 3}}{6} = \frac{\sqrt{9}}{6} = \frac{3}{6} = \frac{1}{2}
\][/tex]
2. Conclusion: As we can see, substituting [tex]\(x = 3\)[/tex] into the function does not result in an indeterminate form (e.g., [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex]). Instead, it gives us a finite value of [tex]\(\frac{1}{2}\)[/tex].
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 3} \frac{\sqrt{2x + 3}}{x + 3} = \frac{1}{2}
\][/tex]
Therefore, the limit of the function as [tex]\(x\)[/tex] approaches 3 is [tex]\(\boxed{\frac{1}{2}}\)[/tex].
