If $f:\mathbb{R}\rightarrow \mathbb{R}$ is differentiable, then $\lim_{h\to 0}\dfrac{f(3+2h)-f(3)}{h}=?$
I wanted to know if you could help me with this exercise, I know that you are using the incremental quotient.
The book's solution:
Here you have to think a little. They are giving us a limit that is very similar to the derivative by definition at point $x_0 = 3$, but it is exactly this derivative, because it would be like this: $$f'(x_0)=\lim_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}\Rightarrow f'(3)=\lim_{h\to 0}\dfrac{f(3+h)-f(3)}{h}$$ As in the limit $f(3)$ appears, we can write it as $kf'(3)$, and that name we give to the exercise limit.
Compare $f'(3)$ and $kf'(3)$ using the definition and the limit they give us: (if they are equal, $k = 1$, and we are left with $f '(3)$, otherwise, it will be some other solution) $$f'(3)=kf'(3)\Rightarrow \lim_{h\to 0}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to 0}\dfrac{f(3+2h)-f(3)}{h}\Rightarrow \lim_{h\to 0}3+h=\lim_{h\to 0}3+2h$$ We have two limits that tend to $0$. We replace and we have left: $$\lim_{h\to 0}3+h=\lim_{h\to 0}3+2h\Rightarrow 3+0=3+2\cdot 0\Rightarrow 3=3 $$ Then, the limits are equal, so $k = 1$. In this way, the answer is $f '(3)$.
Big Hint: $$\frac{f(3+2h)-f(3)}{h}=2\cdot\frac{f(3+2h)-f(3)}{2h},$$ and $h\to 0$ if and only if $2h\to0.$