To solve the limit
[tex]\[
\lim _{x \rightarrow 0} \frac{(3+x)^2-9}{x},
\][/tex]
we can proceed as follows:
1. Expand the numerator:
[tex]\[
(3 + x)^2 = 3^2 + 2 \cdot 3 \cdot x + x^2 = 9 + 6x + x^2.
\][/tex]
2. Subtract 9 from the expanded form:
[tex]\[
(3 + x)^2 - 9 = 9 + 6x + x^2 - 9 = 6x + x^2.
\][/tex]
3. Rewrite the limit using the simplified numerator:
[tex]\[
\frac{(3 + x)^2 - 9}{x} = \frac{6x + x^2}{x}.
\][/tex]
4. Simplify the fraction:
[tex]\[
\frac{6x + x^2}{x} = \frac{x(6 + x)}{x} = 6 + x.
\][/tex]
5. Evaluate the limit as [tex]\( x \)[/tex] approaches 0:
[tex]\[
\lim_{x \rightarrow 0} (6 + x) = 6 + 0 = 6.
\][/tex]
Thus, the limit is
[tex]\[
\lim _{x \rightarrow 0} \frac{(3+x)^2-9}{x} = 6.
\][/tex]
Select the correct answer below:
6
