Consider a differentiable function $$f:\mathbb{R} \rightarrow \mathbb{R}$$. The derivative of the function at any point can be written as: \begin{align*} f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \end{align*} Suppose we have a constant $$c > 0$$, is it true that: \begin{align*} f'(x) = \lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{h} \hspace{3ex}? \end{align*} Since when dividing $$h$$ by some constant in the numerator, it will still become arbitrarily small. Or does it follow that: \begin{align*} \lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{h} = \frac{1}{c}\lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{\frac{h}{c}} = \frac{f'(x)}{c} \hspace{3ex} ? \end{align*} I think that the second case is correct, but I still wanted to be 100% sure.

• It seems to me that you intend to ask two questions (despite the lack of question marks). The answer to the first question is "No", as is easily verifiable by looking at what happens with the identity function. The answer to the second question is yes, the equalities all hold. May 28 '20 at 11:01
• Yes yes, sorry for the confusion. Will edit it May 28 '20 at 11:03

The last equation you wrote is the correct one. You can check with examples. If $$f$$ is the identity, namely $$f(x)=x$$. Then $$\lim_{h\rightarrow 0} \frac{f(x+\frac{h}{c})-f(x)}{h}= \lim_{h\rightarrow 0} \frac{x+\frac{h}{c}-x}{h}=\lim_{h\rightarrow 0} \frac{\frac{h}{c}}{h}=\frac{1}{c}.$$ If $$f(x)=x^2$$ then $$\lim_{h\rightarrow 0} \frac{f(x+\frac{h}{c})-f(x)}{h}= \lim_{h\rightarrow 0} \frac{x^2+2x\frac{h}{c}+\frac{h^2}{c^2}-x^2}{h}=\lim_{h\rightarrow 0} \frac{2x\frac{h}{c}+\frac{h^2}{c^2}}{h}=\lim_{h\rightarrow 0} \frac{2x}{c}+\frac{h}{c^2}=\frac{2x}{c}.$$

• Thanks for the examples! May 28 '20 at 11:06
• You're not exemplifying the last equation. All you are doing is showing examples of what obtains for the penultimate equation. May 28 '20 at 11:12

Here is the proof in case you're interested. It isn't saying much new compared to Anguepa's examples.

$$\begin{array}{} &&\displaystyle \lim_{h \to 0} { f(x + h/c) - f(x) \over h} \\ &=& \displaystyle \lim_{h \to 0} \frac1c { f(x + h/c) - f(x) \over h/c} & \text{...basic algebra}\\ &=& \displaystyle \frac1c \lim_{h \to 0} { f(x + h/c) - f(x) \over h/c} & \text{...pull out the 1/c constant}\\ &=& \displaystyle \frac1c \lim_{cg \to 0} { f(x + g) - f(x) \over g} & \text{...rename g = h/c}\\ &=& \displaystyle \frac1c \lim_{g \to 0} { f(x + g) - f(x) \over g} & \text{...basic property of limit, see below}\\ &=& \displaystyle \frac1c f'(x) &\text{...by definition of f'(x)} \end{array}$$

The only step that actually involves calculus is the $$4$$th equal sign, and further explanation will need to invoke the $$(\epsilon, \delta)$$-based definition of what a limit is. It's a bit tedious, so hopefully you are OK with leaving it as "basic property of limit." :)

• Thanks! That's the most complete answer that I was searching for :). May 31 '20 at 7:51

You have to replace $$h$$ everywhere by $$h/c$$ for the limit to still be $$f'(x).$$ Otherwise, it is generally different.

• So what I did at the bottom was correct? May 28 '20 at 10:59
• Your answer isn't clear at all, as evidenced by P3rs3rk3r's comment above. May 28 '20 at 10:59
• Thanks to you guys! May 28 '20 at 11:03
• @P3rs3rk3r If you mean the last equation in OP then that's incorrect. What you get is still $f'(x).$ May 28 '20 at 11:10
• @DigAmma No, the last equation in OP is incorrect. May 28 '20 at 11:13