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Let $ f $ be a function defined on some neighbourhood of a point $ x_0 $. If the following limit exists $$ \lim_{h\to 0}{\frac{f\left(x_{0}+h\right)-f\left(x_{0}-h\right)}{2h}} $$ then $f$ is said to be symmetrically differentiable at $ x_{0} $ and the above limit is called the symmetric derivative of $ f$ at $ x_{0} $ and is denoted by $ f_{sym}'\left(x_{0}\right) $

Prove that if $ f $ has both right and left derivatives at $ x_0 $ then $ f$ is symmetrically differentiable at $ x_{0} $ (consequently, it would be proved that if $ f$ is differentiable at $ x_{0} $ then $ f$ is symmetrically differentiable at $ x_{0} $)

So I wrote that $$ f_{sym}'\left(0\right)=\lim_{h\to 0}{\frac{f\left(0+h\right)-f(0-h)}{2h}}=\lim_{h\to 0}{\frac{ 0}{2h}} = 0 $$ Is this proof right?

CHAMSI
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NightEye
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  • Welcome to Mathematics Stack Exchange! Are you sure the definition of the symmetric derivative is correct? I think the nominator should be $f(x_0+h)-f(x_0-h)$. – EuklidAlexandria Dec 14 '20 at 17:13
  • What you wrote does not make sense. You define the symmetric derivative to be zero. –  Dec 14 '20 at 17:17
  • Of caurse we can not just replace $ h $ in a part of the expression. When calculating a limit, we replace all the variable at a time. In this case we get an indeterminate form. – CHAMSI Dec 14 '20 at 17:22
  • Try it with $f(x) = x^3$, you do not get 0 – Paul Dec 14 '20 at 17:24

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$$ \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h}=\lim_{h\to 0} \frac{f(x_0 + h)-f(x_0)}{2h}-\lim_{h\to 0} \frac{ f(x_0 - h)-f(x_0) }{2h}$$

Justifies the claim (after you fix it).