Determining $\lim_{h \to 0} \frac{f(x +h) - f(x)}{h}$ for the function $f(x) = 3x^2 + 2x$ Determine $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ for the function $f(x)=3x^2+2x$.
Can anyone please confirm whether this is correct:
\begin{align*}
& = \frac{3x^2+6xh+3h^2+2x+2h-3x^2-2x}{h}\\
& = \frac{6xh+3h^2+2h}{h}\\
& = \frac{h(6x+3h+2)}{h}\\
& = 6x+3h+2
\end{align*}
Do I go a step further by saying:
\begin{align*}
\lim_{h \to 0} (6x+3h+2) & = 6x+(3 \cdot 0)+2\\
                        & = 6x+2
\end{align*}
I am confused about which answer is the final product and if I have gone a step too far.
Thanks for your help.
Apologies for the lack of information earlier.

Given $f(x)=3x^2+2x$:
\begin{align}
\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim\limits_{h\to0}\frac{3(x+h)^2+2(x+h)-2x^2-2x}{h}\\
&=\lim\limits_{h\to0}\frac{6xh+3h^2+2h}{h}\\
&=\lim\limits_{h\to0}\frac{h(6x+3h+2)}{h}\\
&=\lim\limits_{h\to0}\ (6x+3h+2)\\
&=6x+2
\end{align}
You have approached the problem correctly, but it is important to include all necessary notation in your solution (including the $\lim$ in each of your expressions, line after line, until you evaluate the limit by substitution).
