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f'(x)=lim h->0 f(x+h)-f(x)/h, if f(x)= { 2, -10
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With the derivative defined as the limit \[f'(x):=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\]or equivalently (by setting \(h=x-0\), \[f'(0):=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\]in order for the function to be differentiable at \(x=0\), this limit must exist. Given that \[f(x)=\begin{cases}2&\text{if }-10<x<0\\[1ex]1&\text{if }0\le x<10\end{cases}\]the derivative exists at \(x=0\) if \[\lim_{x\to0^-}f'(x)=\lim_{x\to0^-}f'(x)\]Is this the case?
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