Find \( \lim _{h \rightarrow 0} \frac{f(3+h)-I(3)}{h} \) if \( f(x)=x^{2}+8 \) \( \lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\square \) (Simplify your answer.)

To find the limit \[ \lim_{h \rightarrow 0} \frac{f(3+h) - f(3)}{h} \] we start by evaluating the function \( f(x) = x^2 + 8 \) at \( x = 3 \): \[ f(3) = 3^2 + 8 = 9 + 8 = 17. \] Now, we need to find \( f(3+h) \): \[ f(3+h) = (3+h)^2 + 8 = 9 + 6h + h^2 + 8 = 17 + 6h + h^2. \] Now we substitute back into the limit: \[ \lim_{h \rightarrow 0} \frac{f(3+h) - f(3)}{h} = \lim_{h \rightarrow 0} \frac{(17 + 6h + h^2) - 17}{h} \] This simplifies to: \[ \lim_{h \rightarrow 0} \frac{6h + h^2}{h}. \] Next, we can simplify the expression: \[ \lim_{h \rightarrow 0} \frac{h(6 + h)}{h} = \lim_{h \rightarrow 0} (6 + h). \] As \( h \) approaches 0, this limit becomes: \[ 6 + 0 = 6. \] Thus, the final answer is: \[ \lim_{h \rightarrow 0} \frac{f(3+h) - f(3)}{h} = 6. \]