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The definition of the derivative is $$g'(a)=\lim \limits_{\delta \rightarrow 0} \frac{g(a+\delta) - g(a)}{\delta}$$

also the left derivative is $$ \lim \limits_{\delta \rightarrow 0^-} \frac{g(a+\delta) - g(a)}{\delta} \tag{*}$$

and the right derivative is $$ \lim \limits_{\delta \rightarrow 0^+} \frac{g(a+\delta) - g(a)}{\delta} \tag{**}$$

but am I right to think that the left derivative is equivalent to $$ \lim \limits_{\delta \rightarrow 0} \frac{g(a-\delta) - g(a)}{-\delta} = \lim \limits_{\delta \rightarrow 0} \frac{g(a) -g(a-\delta)}{\delta}$$ and the right equivalent to just $$ \lim \limits_{\delta \rightarrow 0} \frac{g(a+\delta) - g(a)}{\delta} $$ if this is not correct then what is the correct equivalent definition of (*) and (**) i.e I want to not have to deal with $0^+$ and want a definition interms of just limit $0$

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  • $\begingroup$ No, if $\delta$ is any real number then your definitions equal simply the usual derivative. $\endgroup$ – user42761 Jun 18 '16 at 10:48
  • $\begingroup$ @Epsilon for both the left and right? $\endgroup$ – Aspire Jun 18 '16 at 10:49
  • $\begingroup$ Yes. What you do is the following: You look at the right derivative, say. This is $(**)$. Then you replace $0^+$ by $0$. But this is simply the derivative at $a$. If the derivative exists, all these numbers are clearly equal. However, there are functions for which the right and left derivative differ. For example $x \mapsto \lvert x \rvert$. $\endgroup$ – user42761 Jun 18 '16 at 10:51
  • $\begingroup$ @Epsilon . So is there a way of writing the definition without the $0^+$ or $0^-$ $\endgroup$ – Aspire Jun 18 '16 at 10:52
  • $\begingroup$ By definition, the left derivative just considers values left from $a$. This means you consider values $a+\delta$, for $\delta < 0$. So $\lim_{\delta \to 0^-}$ means that $\delta$ approaches $0$ from the left. $\endgroup$ – user42761 Jun 18 '16 at 10:55

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