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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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    "Analyze" in this case means "this is equal to something that looks much simpler. Find it."2012-10-21
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    As to what to do, see what happens when you substitute $-h$ for $h$.2012-10-21
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    Thanks so much! So the whole limit goes to 0, because you end up with f' + (-1)f', right?2012-10-21
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    Indeed so. Good +12012-10-21
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    Hope I am not being a bother but if you don't mind I had another question located here: http://math.stackexchange.com/questions/217789/derivative-based-on-continuity primarily about trying to figure out what I am supposed to do. Thanks very much DonAntonio and Gerry, I really appreciate it2012-10-21
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    Not so fast --- take some simple function for $f$, say, $f(x)=x$, evaluate the original limit directly, and see whether you really get zero.2012-10-21
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    Okay will do! Thanks Gerry2012-10-21
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    Hm. When I tried f(x) = x, I ended up getting 2h/h2012-10-21
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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

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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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    I do not see why it is the same other than the fact that the limit worked out identically in each case.2012-10-21
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    Well, $\,x_0+h\,$ and $\,x_0-h\,$ are in the limit when $\,h\to 0\,$ the same as $\,h\,$ approaches zero without any restriction: positive and negative values...if this is what you meant by "the limit worked out identically in each case" the we agree.2012-10-21
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    Anyone care to change their minds in the light of the discussion in the comments?2012-10-21
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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