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I believe I understand this question but I am stuck at what seems to be a "last part."

Here is the question: Suppose that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable at $x_o$. Analyze the following limit: $\lim_{h\ \rightarrow 0} \frac{f(x_o + h) - f(x_o -h)}{h} $.

Analysis:

Observe that $\lim_{h\ \rightarrow 0} \frac{f(x_o + h) - f(x_o -h)}{h} = \lim_{h\ \rightarrow 0} \frac{f(x_o + h) - f(x_o) + f(x_o) -f(x_o -h)}{h} $. Then, applying limit rules, we see that $\lim_{h\ \rightarrow 0} \frac{f(x_o + h) - f(x_o) + f(x_o) -f(x_o -h)}{h} = \lim_{h\ \rightarrow 0} \frac{f(x_o + h) - f(x_o)}{h} + \lim_{h\ \rightarrow 0} \frac{f(x_o) -f(x_o -h)}{h} = f'(x_o) + \lim_{h\ \rightarrow 0} \frac{f(x_o) -f(x_o -h)}{h}$

It is here that I am stuck. How do I deal with that right-most limit directly above, after the "plus"? Also, is this what was desired in terms of "analysis" ?

thanks

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    The second limit is $\lim_{h\to 0} -\frac{f(x_0)-f(x_0+h)}{h}$. If you'll notice, the numerator is reversed: so the negatives cancel, and you in fact get $2f'(x)$ as your final result. If you originally divided by $2h$, this would be the symmetric derivative.2012-10-21

1 Answers 1

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Hint:

$\frac{f(x_0+h)-f(x_0-h)}{h}=\frac{f(x_0+h)-f(x_0)}{h}-\frac{f(x_0-h)-f(x_0)}{h}$

And now just be sure you understand why in the definition of derivative it is the same to

have $\,f(x_0+h)\,$ or to have $\,f(x_0-h)\,$ in the numerator...

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    Actually, I think it would have been better to write $\frac{f(x_0+h)-f(x_0-h)}{h}=\frac{f(x_0+h)-f(x_0)}{h}+\frac{f(x_0-h)-f(x_0)}{-h} = 2f'(x_0)$2015-07-08