# A basic question on the derivative of a continuous function

Is the following condition necessary for the existence of derivative of a continuous function at point $x$:

$$\lim_{h \to 0^+}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0^-}\frac{f(x)-f(x+h)}{h}$$

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If the derivative of $f$ exists, then the left-hand side there will be $f'(x)$ and the right-hand side will be $-f'(x)$. If you fix this by putting a minus sign on the right, then your statement is a necessary and sufficient condition for the derivative to exist, since it's just the definition of the derivative combined with the fact that a limit exists iff it exists from each side and the one-sided limits are equal.

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No. In the current form, the condition says the derivative from the left and from the right have opposing signs. For a simple counter-example, consider the function $f(x)=x$ at any point; the left-hand limit will be equal to 1, while the right-hand one would be equal to (-1).

If reverse the order of $f(x+h)$ and $f(x)$ in the limit on the right-hand side, the condition would be both necessary and sufficient since it'd basically be the definition of a derivative (in order for a limit in $0$ to exist, it must be the same regardless of the way you approach $0$).

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If you change the denominator of the second limit to $-h$ then yes but as it stands that's actually not true in general (think of the signs of the numerator and denominator)

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No. The readily available counterexample is $f(x)=x$ which though differentiable at $0,$

$$\displaystyle\lim_{h\to0+}\frac{f(h)}{h}=1\\\displaystyle\lim_{h\to0-}\frac{f(h)}{-h}=-1$$

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