Let $f(t)$ be a Lipschitz continuous function and let $ f'(t)=\lim_{\delta\downarrow 0 } \frac{f(t+\delta)-f(t)}{\delta}$ denote its one-sided right derivative in $t$. Assume that $f'(0)$ exists. My question is about the equation: $$ f'(0) = \lim_{t\downarrow 0} f'(t). $$ Is this equation necessarily true provided that the limit in the RHS exists?
In general, the equation above does not make sense because the RHS may not exists even when $f'(0)$ does, for instance $f(x)=x^2\sin(1/x)$, but if also the RHS exists, does it need to be equal to $f'(0)$?
Since $f$ is Lipschitz continuous, the both side derivative exists almost everywhere and the fundamental theorem of calculus applies: $$ f(t) - f(0) = \int_0^t f’(x) dx. $$
Let $a = \lim_{x \downarrow 0} f’(x)$. For $\epsilon > 0$ there exists some $\delta > 0$ such that for all $0< x < \delta$ it follows $$ |f’(x) - a| < \epsilon. $$ Thus, we have $$ \left | \frac{f(t) - f(0)}{t - 0} - a\right| \le \frac1t \int_0^t |f’(x) -a | \le \frac t t \epsilon. $$ That is, the one-side derivative $f’(0)$ exists and is equal $a$.