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what's the derivative of $f(x)= x^{2}$ ($x\geq 0$) when x=0? from my understand, it doesn't exist because even $\lim_{h \to 0^{+}}\frac{f(x+h)-f(x)}{h}$ is 0, but $\lim_{h \to 0^{-}}\frac{f(x+h)-f(x)}{h}$ doesn't exist, am I correct?

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Short answer: You are incorrect because when calculating a limit of a function, you only need to look at its domain. So, if a function is defined on $[a,b]$, then its limit at $a$ is equal to its right limit at $a$.

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  • $\begingroup$ It's not a limit of a function. don't we need to make right hand limit equal to left hand limit to say the limit exist? $\endgroup$ – whoisit Feb 23 '16 at 10:18
  • $\begingroup$ @whoisit Of course it is the limit of a function. It is the limit of the function $g_x: h\mapsto \frac{f(x+h) - f(x)}{h}$, and if $f$ is defined on $[a,b]$, then the function $g_a$ is defined for $h>0$. $\endgroup$ – 5xum Feb 23 '16 at 10:19
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If you look for the rigorous definition of limit, then you may understand that $f'(0)$ exists. In high-school or calculus course, we usually learn limit naively because rigorous approach is too hard. But sometimes it occurs cognitive obstacles, and thinking that limit exists only if left and right limit must exist like you is also cognitive obstacles. Let's introduce rigorous definition of limit of function- Let $A\subset \mathbb{R}$ be a domain of $f$.

If for every $\epsilon > 0$, there exists $\delta > 0$ such that if for all $x\in A$, $|x-a|<\delta$ then $|f(x)-L|<\epsilon$, then we say $f$ converges to $L$ as $x\to a$.

Then how about left or right sided limit?

If for every $\epsilon > 0$, there exists $\delta > 0$ such that if for all $x\in A$, $a < x <a+\delta\;(a-\delta < x < a)$ then $|f(x)-L|<\epsilon$, then we say that $L$ is a right(left) sided limit of $f$.

How about $f(x)=x^2\;(x\ge 0)$? Here $A=[0,\infty)$, then if $x\in (-\delta,0)\cup(0,\delta)$ and $x\in A$, then $A\cap (-\delta,\delta)=(0,\delta)$, thus $x\in (0,\delta)$. Thus right sided limit is enough.

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