Certainly! Let's evaluate the limit of the function [tex]\(\frac{\sqrt{x+8}}{\sqrt{x}}\)[/tex] as [tex]\(x\)[/tex] approaches 1.
Given:
[tex]\[ \lim_{x \to 1} \frac{\sqrt{x+8}}{\sqrt{x}} \][/tex]
### Step-by-Step Solution:
1. Substitute [tex]\(x = 1\)[/tex]
Start by attempting a direct substitution to see if we encounter any indeterminate forms. Substitute [tex]\( x = 1 \)[/tex] in the expression:
[tex]\[
\frac{\sqrt{1+8}}{\sqrt{1}} = \frac{\sqrt{9}}{\sqrt{1}} = \frac{3}{1} = 3
\][/tex]
Since substituting [tex]\( x = 1 \)[/tex] into the expression does not result in an indeterminate form, the limit can be directly evaluated:
[tex]\[ \boxed{3} \][/tex]
Therefore,
[tex]\[ \lim_{x \rightarrow 1} \frac{\sqrt{x+8}}{\sqrt{x}} = 3 \][/tex]
