Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I came across this:

If $\lim\limits_{h\to0}\frac{f\left( 1+h \right)}{h}=1$ and $f(x)$ has a derivative at $x=1$ then: $$\lim_{h\to0}\frac{f\left( 1 \right)}{h}=\lim_{h\to0}\frac{f\left( 1+h \right)}{h}-\lim_{h\to0}\frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$$

I don't understand why this is true. Can someone explain it to me?

Thanks :)

share|cite|improve this question
@Jason : You had some complicated code involving \underset and \mathop where you only needed \lim_{h\to 0}. I changed it. In the one instance where it's "inline" rather than "displayed", I changed it to \lim\limits_{h\to 0}. – Michael Hardy Dec 16 '11 at 19:15
@MichaelHardy Ok, Thanks :) I don't know TeX I just use a program to generate it - I guess that's why it's messy sometimes – Jason Dec 16 '11 at 19:16
up vote 3 down vote accepted

This is a consequence of one of the basic limit laws:

If $\lim\limits_{t\to a} F(t)$ exists and $\lim\limits_{t\to a}G(t)$ exists, then $\lim\limits_{t\to a}\Bigl( F(t)-G(t)\Bigr)$ exists, and $$\lim_{t\to a}\Bigl(F(t)-G(t)\Bigr) = \lim_{t\to a}F(t) - \lim_{t\to a}G(t).$$

So, we are assuming that $$\lim_{h\to 0}\frac{f(1+h)}{h}\ \text{exists.}$$ We are also assuming that $f(x)$ has a derivative at $1$; that means that $$\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\ \text{exists.}$$

Therefore, $$\lim_{h\to 0}\left( \frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right)\ \text{exists}$$ and moreover, $$\lim_{h\to 0}\left(\frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right) = \lim_{h\to 0}\frac{f(1+h)}{h} - \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$$ by the limit law quoted above, with $F(h) = \frac{f(1+h)}{h}$ and $G(h) = \frac{f(1+h)-f(1)}{h}$.

Now just notice that $$\begin{align*} \frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h} &= \frac{f(1+h)-(f(1+h)-f(1))}{h}\\ &= \frac{f(1+h)-f(1+h)+f(1)}{h}\\ &= \frac{f(1)}{h}, \end{align*}$$ giving the equality you have when you put everything together.

That means that $$\begin{align*} \lim_{h\to 0}\frac{f(1)}{h} &= \lim_{h\to 0}\left(\frac{f(1+h)}{h} - \frac{f(1+h)-f(1)}{h}\right)\\ &= \left(\lim_{h\to 0}\frac{f(1+h)}{h}\right) - \left(\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\right)\\ &= 1 - f'(1), \end{align*}$$ since your assumption is that the first limit equals $1$, and by definition the second limit is the derivative at $1$.

share|cite|improve this answer
Thanks :) Great answer – Jason Dec 16 '11 at 19:32

It doesn't look as a useful trait but it is true. We have: $$\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1 \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\ \frac{f \left( 1+h \right) - f\left( 1+h \right)+f\left( 1 \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\ \frac{f \left( 1+h \right) - (f\left( 1+h \right)-f\left( 1 \right))}{h}=$$$$=\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1+h \right)}{h}-\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$$

share|cite|improve this answer
Thank you :) I agree I think it not useful very often but it is for a particular problem I am trying to solve... – Jason Dec 16 '11 at 19:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.