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I am reading a text where directional derivatives of functions $f:E\rightarrow F$, where $E,F$ are Banach spaces, at the point $x_0 \in E$ are defined as $d_v f(x_0)=\lim_{t\rightarrow 0+} \frac{f(x_0+tv)-f(x_0)}{t}$ for any $v\in E$. My question is: Why is the limit taken with respect to "$t\rightarrow 0+$" instead of just "$t\rightarrow 0$" ? What does it change if we have "$t\rightarrow 0+$ instead of "$t\rightarrow 0$" ? Since in some textbooks, just the ""$t\rightarrow 0$" version is used.

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Some functions will not be derivable if you take $t\to 0$. For instance the absolute value function on $\mathbb{R}$. So I suppose it all comes down to what they want to do with this derivative. – Raskolnikov Jun 9 '11 at 13:15
I have never seen the directional derivative defined in this manner; can you provide the name / page of the text? – ItsNotObvious Jun 9 '11 at 13:18
I am confused about defining ratios in general Banach spaces. Shouldn't there be a norm there somewhere? – gary Jun 9 '11 at 14:00
I guess they want to be able find directional derivatives of points on the boundaries of the domain of the function. – Dactyl Jun 9 '11 at 14:09
If you are lucky the $\lim_{t\to 0}$ is linear in $v$ while for the $\lim_{t\to 0+}$ this is not an issue. – Christian Blatter Jun 9 '11 at 15:10
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