A limited problem

For any $x>0$ and $k>0$ , $\frac{k}{x}-1<\left \lfloor \frac{k}{x} \right \rfloor \le \frac{k}{x}$

Multiplying through by $x$ , $k-x<x\left \lfloor \frac{k}{x} \right \rfloor \le k$

Now we can use the squeeze theorem; as $x \to 0$ , $x\left \lfloor \frac{k}{x} \right \rfloor \to k$ .

Applying this to the sum in the question, the limit is $1+2+\cdots+100=T_{100}=\boxed{5050}$

Note that:

$\left\lfloor\frac{a}{x}\right\rfloor=\frac{a}{x}-\left\{\frac{a}{x}\right\}$ Where $\left\{\dfrac{a}{x}\right\}$ is the fractionary part of $\dfrac{a}{x}$

$\implies \lim_{x\rightarrow 0}x\left[\left(\frac{1}{x}+\frac{2}{x}+...+\frac{100}{x}\right)-\left(\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+...+\left\{\frac{100}{x}\right\}\right)\right]$ $\lim_{x\rightarrow 0}\left(1+2+3+...+100\right)-\left(x\left\{\frac{1}{x}\right\}+x\left\{\frac{2}{x}\right\}+...+x\left\{\frac{100}{x}\right\}\right)$

Now, there are two cases. If $\left(\dfrac{a}{x}\right)$ is an integer, then $\left\{\dfrac{a}{x}\right\}=0$ and consecuently $x\left\{\dfrac{a}{x}\right\}=0$ .

If $\dfrac{a}{x}$ is not an integer, then $0<\left\{\dfrac{a}{x}\right\}<1$ , $\implies0<x\left\{\dfrac{a}{x}\right\}<x$

In the second case as $x\rightarrow 0$ $\implies x\left\{\dfrac{a}{x}\right\}\rightarrow 0$ . It gets squeezed to 0.

So all in all, the limit will just be the sum of the integers from 1 to 100 , as the other terms go to 0.

Since $x \rightarrow 0$ , the denominator is infinitesimally small and hence the quotient approaches closer and closer to the numerator.

So it can be rewritten as $1+ 2+ 3+ .... + 100$

Which will be equal to $50× 101 = 5050$