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I see in this Wolfram article that Riemann integral is defined as $$\lim_{\max \Delta x_k \to 0} \;\sum_{k=1}^n f\!\left(x^*_k\right)\,\Delta x_k$$

Since $\forall k \in \mathbb{N}^+ : \Delta x_k > 0$, should not the definition uses a one-sided limit as the following one instead? $$\lim_{\max \Delta x_k \to 0^+} \;\sum_{k=1}^n f\!\left(x^*_k\right)\,\Delta x_k$$

Tadeus Prastowo
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Let me answer that question with another example, and I hope it clears things up.

We write $$\lim_{x\to 0} \sqrt{x}=0$$ although $$\lim_{x\to 0^{+}} \sqrt{x}=0$$

would also be okay. The first you will see more often.

If you want me to be more explicit:

When discussing limits, we only care about the domain of the function. In this case $\Delta x_k>0$. So it really IS a two sided limit. It's just that there is no "left side" to consider.

  • Is it similar to the case where in a definition people often use "if" when they really mean "if and only if", for example, "a function $f$ is differentiable at a point $c$ if $\lim_{x\to c} \frac{f(x) - f(c)}{x - c}$ exists"? – Tadeus Prastowo Nov 24 '15 at 06:22
  • @TadeusPrastowo A definition is ALWAYS an if and only if –  Nov 24 '15 at 06:23
  • Yes, it is. And, based on that understanding, people can be a bit sloppy and write "if" in place of "if and only if", which is something that cannot be done when stating a theorem. So, based on your answer, I can conclude that when the domain is ${x | x \geq 0}$, people can also be a bit sloppy and write $\lim_{x\to 0}$ in place of $\lim_{x\to 0^+}$, can't I? – Tadeus Prastowo Nov 24 '15 at 06:42
  • @TadeusPrastowo There is NOTHING sloppy about writing $\lim_{x\to 0}$ when the domain is ${x \mid x \geq 0}$. It is completely 100% okay. –  Nov 24 '15 at 06:45
  • @TadeusPrastowo Perhaps examining the formal definition of a limit would be helpful. There is a definition on Wikipedia https://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit#Precise_statement –  Nov 24 '15 at 06:48
  • Suppose that is not sloppy. Then, $\lim_{x\to 0} \sqrt{x} = 0 = \lim_{x\to 0^+} \sqrt{x} = \lim_{x\to 0^-} \sqrt{x}$ is a must. By $\varepsilon$-$\delta$ definition of left-hand limit: $\lim_{x\to 0^-} \sqrt{x} = 0 \Longleftrightarrow \forall \varepsilon\in\mathbb{R} : \exists \delta\in\mathbb{R} : \forall x \in \emptyset : -\delta < x < 0 \longrightarrow \left|\sqrt{x} - 0\right| < \varepsilon$ is true because $\forall x \in \emptyset$ is trivially true, and therefore, it is not sloppy to write $\lim_{x\to 0}$ when the domain is ${x | x \geq 0}$, is it? – Tadeus Prastowo Nov 24 '15 at 07:03
  • @TadeusPrastowo That is the idea. There are some technical concerns about accumulation points I'd have to think about more, but you have the right idea. –  Nov 24 '15 at 07:05