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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 $f '(3)$.

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Jul 29 '18 at 16:28
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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 $2h\to0.$

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  • $\begingroup$ Thanks! So, this limit is equal to $2\cdot f'(3)$, right? $\endgroup$ – a5537539 Jul 29 '18 at 16:31
  • $\begingroup$ In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved. $\endgroup$ – a5537539 Jul 29 '18 at 16:37
  • $\begingroup$ Bingo! Can you justify it? $\endgroup$ – Cameron Buie Jul 29 '18 at 16:37
  • $\begingroup$ Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct. $\endgroup$ – Cameron Buie Jul 29 '18 at 16:38
  • $\begingroup$ Then I do $t=2h$ and $h\to 0$ if and only if $t\to 0$, and I will have $2\lim_{t\to 0}\dfrac{f(3+t)-f(3)}{t}=2f'(3)$ $\endgroup$ – a5537539 Jul 29 '18 at 16:38
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Hint

What is $$\lim\limits_{h \to 0} \frac{f(3+2h)-f(3)}{2h}?$$

What can you conclude from this limit?

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  • $\begingroup$ In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved. $\endgroup$ – a5537539 Jul 29 '18 at 16:37
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$$\lim_{h\to 0}\dfrac{f(3+2h)-f(3)}{h}= \lim_{2h\to 0} 2\dfrac{f(3+2h)-f(3)}{2h} = 2f'(3) $$

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