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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 :)

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@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
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2 Answers 2

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 $1$.

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Thanks :) Great answer –  Jason Dec 16 '11 at 19:32
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It doesn't look as a useful trait but it is true. We have: $$\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1 \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\ \frac{f \left( 1+h \right) - f\left( 1+h \right)+f\left( 1 \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\ \frac{f \left( 1+h \right) - (f\left( 1+h \right)-f\left( 1 \right))}{h}=$$$$=\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1+h \right)}{h}-\underset{h\to 0}{\mathop{\lim }}\ \frac{f\left( 1+h \right)-f\left( 1 \right)}{h}$$

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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
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