4. $limlimits_{x \to 1}\frac{sqrt{x + 8}-3}{x - 1}$

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + 8}+3$. $$\lim_{x\rightarrow1}\frac{\sqrt{x + 8}-3}{x - 1}=\lim_{x\rightarrow1}\frac{(\sqrt{x + 8}-3)(\sqrt{x + 8}+3)}{(x - 1)(\sqrt{x + 8}+3)}$$ Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(x + 8)-9=x - 1$. So the limit is $\lim_{x\rightarrow1}\frac{x - 1}{(x - 1)(\sqrt{x + 8}+3)}$.

Step2: Simplify the expression

Cancel out the common factor $(x - 1)$ (since $x\neq1$ as $x\rightarrow1$). We get $\lim_{x\rightarrow1}\frac{1}{\sqrt{x + 8}+3}$.

Step3: Substitute $x = 1$

Substitute $x = 1$ into $\frac{1}{\sqrt{x + 8}+3}$. When $x = 1$, $\sqrt{1+8}+3=\sqrt{9}+3=3 + 3=6$. So the limit is $\frac{1}{6}$.

Answer:

$\frac{1}{6}$