# Analysis of a limit

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

-
"Analyze" in this case means "this is equal to something that looks much simpler. Find it." – Gerry Myerson Oct 21 '12 at 5:04
As to what to do, see what happens when you substitute $-h$ for $h$. – Gerry Myerson Oct 21 '12 at 5:05
Thanks so much! So the whole limit goes to 0, because you end up with f' + (-1)f', right? – Arthur Collé Oct 21 '12 at 5:09
Indeed so. Good +1 – DonAntonio Oct 21 '12 at 5:15
Hope I am not being a bother but if you don't mind I had another question located here: math.stackexchange.com/questions/217789/… primarily about trying to figure out what I am supposed to do. Thanks very much DonAntonio and Gerry, I really appreciate it – Arthur Collé Oct 21 '12 at 5:19

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...

-
I do not see why it is the same other than the fact that the limit worked out identically in each case. – Arthur Collé Oct 21 '12 at 5:17
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. – DonAntonio Oct 21 '12 at 5:21
Anyone care to change their minds in the light of the discussion in the comments? – Gerry Myerson Oct 21 '12 at 11:56
Actually, I think it would have been better to write $\frac{f(x_0+h)-f(x_0-h)}{h}=\frac{f(x_0+h)-f(x_0)}{h}+\frac{f(x_0-h)-f(x_0)}{-h‌​} = 2f'(x_0)$ – Steven Gregory Jul 8 '15 at 5:55