This question is probably meaningless. I am the author. Please delete it. Thank you.
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1for what stands $$f(x_i)$$? – Dr. Sonnhard Graubner Apr 01 '17 at 06:29
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It is the value of a function at every $x_i$, where $i$ is described by the summation. – Math12345 Apr 01 '17 at 06:32
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1Your question is wholly dependent on $f$. I seriously doubt there is a closed form for arbitrary $f$. – Clayton Apr 01 '17 at 06:35
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Ok. I will modify my question. Thanks. – Math12345 Apr 01 '17 at 06:37
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@Math12345 If you put the limit above sigma a and the right b, you may know both has a difference. – Takahiro Waki Apr 01 '17 at 07:33
1 Answers
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In your question as originally stated you were asking about
\begin{equation} \lim_{h\to0}\sum_{i=1}^{\frac{b-a}{h}}f(x_i)h\tag{1} \end{equation}
This is a Riemann sum which, if $f$ is continuous on the interval $[a,b]$ approaches
\begin{equation} \int_a^bf(x)\,dx \end{equation}
In the expression the interval $[a,b]$ is subdivided into $n>0$ subintervals of equal length $h=\dfrac{b-a}{n}$ which means that $n=\dfrac{b-a}{h}$. Therefore equation $(1)$ can be written
\begin{equation} \lim_{h\to0}\sum_{i=1}^{n}f(x_i)h\tag{2} \end{equation}
On each subinterval an $x_i$ is chosen and a rectangle is formed with base the $i$th subinterval of $[a,b]$ which has length $h$ and height equal to $f(x_i)$. Thus the area of the $i$th rectangle is $f(x_i)h$. The sum of all the rectangles is the summation in equation $(2)$.
Now as $n\to\infty$ we have $h\to0^+$ so the sum of the rectangles will approach the area under $f$ on $[a,b]$ as a limit. Notice that $h$ is always positive so $h\to0^-$ makes no sense in this context, so it is actually
\begin{equation} \lim_{h\to0^+}\sum_{i=1}^{n}f(x_i)h\tag{2}=\int_a^bf(x)\,dx \end{equation}
This is why Wolfram said (according to your original statement of the problem) that the two-sided limit does not exist.
John Wayland Bales
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