How do I evalute this?

Question

This question is probably meaningless. I am the author. Please delete it. Thank you.

Answer 1

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.