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Suppose that $f$ is a real function and that $f'(a)$ exists:

$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. $$

By replacing all instances of $h$ with $-h$, I can get an equivalent definition

$$ f'(a) = \lim_{-h \to 0} \frac{f(a-h) - f(a)}{-h}. $$

Question:

Is it right to say that I can replace $\lim_{-h \to 0}$ with $\lim_{h \to 0}$ in the second definition because $-h$ tending to $0$ is the same as saying $h$ tends to $0$ from the left/bottom?

Since $f'(a)$ exists, the left limit must be equal to the right limit and hence the switch is valid?

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Well remember that for $\lim\limits_{h\to 0}$ to exist, both $\lim\limits_{h\to 0^+}$ and $\lim\limits_{h\to 0^-}$ should both exist and be equal. With that, both your forward difference and backward differences will exist if the corresponding directional derivatives exist... –  J. M. Aug 14 '12 at 11:24
    
well, $\lim_{h \rightarrow 0}$ normally means you allow $h$ to approach 0 from either side (i.e. both tiny positive numbers and tiny negative numbers), so yeah, same thing. –  uncookedfalcon Aug 14 '12 at 11:24
    
Thank you so much for the help! –  Legendre Aug 14 '12 at 11:26
    
I've up-voted Brian M. Scott's answer, but I think mine also covers it and is simpler than his. See below$\ldots\,{}$. –  Michael Hardy Aug 14 '12 at 17:00
    
Hmmm...sorry if this wasn't clear in my question: I was trying to replace only the (-h -> 0) in the limit notation, instead of replacing both (-h -> 0) in the limit notation and (-h) in the fraction. –  Legendre Aug 15 '12 at 16:18
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2 Answers 2

up vote 4 down vote accepted

Neither statement involves a one-sided limit. In both statements $h$, whether as $h$ or as $-h$, can be either positive or negative.

Both statements are shorthand. The first says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a+h)-f(a)}h-f'(a)\right|<\epsilon\tag{1}$$ whenever $0<|h|<\delta$. The second says that for each $\epsilon>0$ there is a $\delta>0$ such that $$\left|\frac{f(a-h)-f(a)}{-h}-f'(a)\right|<\epsilon\tag{2}$$ whenever $0<|-h|<\delta$. Now $|-h|=|h|$, so in both cases you're simply saying that a certain inequality involving $f'(a)$ is true whenever $-\delta<h<0$ or $0<h<\delta$. There is no implication that $h$ is positive or that $-h$ is negative. The question is whether saying that $(1)$ is true whenever $-\delta<h<0$ or $0<h<\delta$ is the same thing as saying that $(2)$ is true whenever $-\delta<h<0$ or $0<h<\delta$.

The answer is yes. Let $$q_1(h)=\frac{f(a+h)-f(a)}h$$ and $$q_2(h)=\frac{f(a-h)-f(a)}{-h}\;.$$

A little algebra shows that $q_1(-h)=q_2(h)$ for all non-zero $h$. As $h$ takes on all non-zero values between $-\delta$ and $\delta$, so does $-h$; it just does so in reverse order, so to speak. Thus, saying (as in $(1)$) that $|q_1(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$ is exactly the same as saying that $|q_1(-h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$, which is the same as saying (as in $(2)$) that $|q_2(h)-f'(a)|<\epsilon$ whenever $-\delta<h<0$ or $0<h<\delta$.

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You have $\lim\limits_{h\to0}\cdots\cdots$.

Let $j=-h$. Notice that as $h\to0$, you also get $j\to0$. Since $h=-j$, you can put $-j$ wherever you had $h$, and $j\to0$ where you had $h\to0$.

Then, since it doesn't matter by which name you call a variable, you can call it $h$ instead of $j$. And there you have it.

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