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When working through an analysis problem involving derivaties I have found myself at this limit :

$\lim_{h \to 0} \frac{g(x_0)-g(x_0-h)}{h}$
Doing some testing with well-defined functions leads me to believe this may be some sort of alternative definition of a derivative, can anyone shed some light?

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    Replace $h$ by $-h$. – John Douma Jan 29 '22 at 22:18
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    In terms of solving the limit, am I allowed to just replace $h$ with $-h$? –  Jan 29 '22 at 22:23
  • this is the definition for the 'left side derivative' at point $x_0$ EDIT: nevermind, this is only the case for $h>0 $ – Danny Blozrov Jan 29 '22 at 22:29
  • In the definition of the limit $h$ lies in a neighborhood of 0 (that doesn't necessarily include 0) i.e. $0<|h|<\delta.$ $h$ can be positive or negative, and we should have no problem replacing $h$ with $-h$ in this limit. – user317176 Jan 29 '22 at 22:35
  • You posted the same question less than an hour before you posted this one: https://math.stackexchange.com/questions/4369295/differentiability-in-relation-to-strange-limit Please don't do this... – Adam Rubinson Jan 29 '22 at 23:17

2 Answers2

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Let $t=-h$. As $h$ tends to zero $t$ does so as well. Therefore the limit is equivalent to $$\lim_{t\to 0} \frac{g(x_0)-g(x_0+t)}{-t} = \lim_{t\to 0} \frac{g(x_0+t)-g(x_0)}{t}$$ and if this limit exists it equals $g'(x_0)$.

Snaw
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Geometrically this represents (in the derivative) that you are not starting at the point $x_0$ and taking a little increment t, indeed, you are parting at the end of the increment (t) and you are form this start point to the point $x_0$ (the narrow reverse its sign) and then, when you do t tends to zero, the star point of this narrow is reaching more and more and more to the point $x_0$

Hidden Arrows
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