Why isn't $\lim_{x \to 0^+} x \left[ \frac px \right] = 0$? 
Find $L$ 
  $$\lim_{x\to{0^+}}  x\left(\left[\frac 1x\right] + \left[\frac 2x\right] + \left[\frac 3x\right] + \cdots + \left[\frac {12}x\right]\right) = L$$
  Here $[t]$ represents the greatest integer less than or equal to $t$.  

Now since $$\lim_{x \to 0^+} x \left[ \dfrac px \right] = \lim_{x \to 0^+} x\left( \frac px - \left\{\frac p x\right\}\right) = \lim_{x \to 0^+} \left(p-x \left\{\frac px\right\}\right) = p$$  
I get $L=1+2+3 +\cdots + 12 = 120$. This would seem fine.  
Here's what confuses me: 
$x$ is a really small positive number. So $\frac 1x$ would be an incredibly large number. So $[\frac 1x]$ would also be large. Now we multiply it with $x$ which is effectively zero. So $x \times [\frac 1x]$ should be zero. And subsequently $L$ should be zero.  
So it boils down to the following:
Why isn't $\lim_{x \to 0^+} x \left[ \dfrac px \right] = 0$ ? When all we're doing is multiplying zero with a very large number to yield zero.  
All help is appreciated! 
 A: 
When all we're doing is multiplying zero with a very large number to yield zero.

No, that's not what we're doing.
$x$ is not zero in that expression, any more than ${p\over x}$ is $\infty$ (let's ignore the "$[]$" for now). If we imagine setting $x$ to be literally zero, then we would looking at $$0\cdot {p\over 0}=0\cdot\infty.$$ Note how the "$p\over x$" part also did something extreme, with the result that the "value" we get is completely meaningless. In general, we can't compute $\lim_{x\rightarrow a}f(x)$ by just thinking about $f(a)$, and this is an example of that.
Your confusion arises because you're thinking of the first $x$ in "$x\cdot [{p\over x}]$" as actually being $0$ but you're not thinking of the second $x$ as actually being $0$. Indeed, neither of the $x$s is actually zero: when evaluating the expression "$\lim_{x\rightarrow0^+}x\cdot[{p\over x}]$," we're looking at positive but nonzero $x$ (look back at the precise definition of a limit).
A: Your "So $\dfrac 1x$ would be an incredibly large number" is contradictory with "we multiply it with $x$ which is effectively zero".
If $\dfrac 1x$ would be an incredibly large number, then $x$ would be an incredibly small number, and their product... $1$.
