It is possible, null already exists.


NULL

• Null is a special value for a pointer (or other kind of object reference) used to signify that the pointer intentionally does not have a target. Such a pointer is called a null pointer. Many implementations use a binary 0 (all bits zero) as the null value, as most operating systems consider it an error to try to access such a low memory address. Some languages use other nomenclature for such a pointer, e.g., nil, undefined, void reference, etc.

• In many disciplines, the concept of null allows a three-valued logic, with null indicating "unknown value". The SQL database query language uses Null in this way, as do Visual Basic (termed "Nothing") and its derivatives. In this model, an expression that depends on the value of a Null operand will evaluate to Null (VB) or "unknown" (SQL). So, for any A, the expressions "A = Null" and "A <> Null" are neither true nor false. However, the boolean operation "A and False" produces false, and similarly "A or True" is true, even when A is Null, because these expressions do not depend on the value of A. (Some SQL implementations may consider A = Null to be "true" if A is Null; see Null (SQL)).

http://en.wikipedia.org/wiki/Null_%28computer%29

 

Null = my interest in this thread.

Beg = wanting something....like my minute back.

 

if you're so smart, find the lim x->0 of x/x

 

1?

 

uhh ok i'm lost on that one lama guy...

google said 1 tho

 

The area of the circular wedge is 1/2 x while the triangle ABC
has area 1/2 sin(x) and the triangle ABD has area 1/2 tan(x). From
the diagram, clearly for 0 < x < pi/2, sin(x) < x < tan(x). Thus,
dividing by x, we get
sin(x) tan(x)
------ < 1 < ------ [1]
x x

multiplying [1] by cos(x) yields

sin(x) cos(x) sin(x)
------------- < cos(x) < ------ [2]
x x

Combining [1] and [2], we have for 0 < x < pi/2,

sin(x)
cos(x) < ------ < 1 [3]
x

Each term in [3] is even, so [3] holds for -pi/2 < x < 0 as well.
Therefore, [3] holds for -pi/2 < x < pi/2, except x = 0.

As x -> 0, cos(x) -> 1. Since sin(x)/x is sandwiched between cos(x)
and 1, we must have

sin(x)
lim ------ = 1
x->0 x

 

I'll tell you tomorrow when I start studying for my calc final.