tanning

Factual Questions
1

Alright…

tan 90 = undefined
cot 90 = 0

so 1/undefined=0

Is this always true?

2

No. 1/undefined = undefined. The result of any operation involving an undefined value is undefined.

3

Nope. One might say never.

tan x = sin x / cos x
cot x = cos x / sin x

tan x is undefined when cos x = 0, i.e., x = 90 + 180k, k in the set of integers.
cot x is undefined when sin x = 0, i.e., x = 180k, k in the set of integers.

4

Dammit. And here I was getting all ready to give a lecture on the dangers of excessive UV light exposure.

5

Well i crammed cot x into a graphing calculator and I got a y value of 0 when x was 90.

The way I figure it

cot 90 = 0
1/tan 90 = 0
tan 90 = undefined
1/undefined = 0

what’s wrong with this logic?

6

The error is that not all mathematical operations are reversable. Dividing any number by zero gives the undefined value. Any calculation involving the undefined value, by definition, gives the undefined value.

1/undefined = undefined
1 + undefined = undefined
undefined[sup]3[/sup] = undefined
undefined * 3.14159265358979323844 = undefined

Capisce?

7

This is actually a battle over semantics. “Undefined” as used here means Infinity. There are Undefined values that aren’t so simple (0/0, for example), but this one is fairly simple.

What you’re really saying is this:


Lim        sin(x)
x->pi/2    ------ = infinity.
           cos(x)

In english, no matter how close you get to pi/2, as long as you’re not actually THERE, you have a real value for |tan(x)| (I don’t want to bother with negatives) that can be calculated. And the closer you get, the bigger that value gets. Effectively (although not technically), dividing by zero gives you infinity. Try it: Divide a cake into Infinity equal pieces. How big is each piece? ZERO.

In the same sense, dividing by infinity gives an effective result of zero. This is used in an application of L’Hopital’s rule:


If f(c)=0 & g(c)=0, then:
Lim    f(x)   f'(x)   f"(x)
x->c   ---- = ----- = -----
       g(x)   g'(x)   g"(x)

BUT, If f(c)=undefined (division by zero) and g(c)=undefined, it still works, because:
Lim   f(x)   1/g(x)           1/(a/0)   0/a   0
x->c  ---- = ------  because  ------- = --- = -
      g(x)   1/f(x)           1/(b/0)   0/b   0
8

I’m sorry, but I can’t help but throw in my two cents here. I have absolutly NO clue as to what any of you folks are talking about. For all I know you could be discussing the dimensions of a solar system or some new formula on splitting atoms. My only reason for responding to this, is to represent the common idiot. I would also like to congratulate all of you who seem to have exceled at a level of mathematics that I can only imagine. So in saying that, Bravo! and keep up the number crunching.

9

Quibbles:


   lim     sin(x)
x -> pi/2+ ------ =  infinity
           cos(x)
---------------------------------
   lim     sin(x)
x -> pi/2- ------ = -infinity
           cos(x)

So, the limit is not infinity, but truly undefined.

L’Hôpital’s rule, as you state, is used when both the top and the bottom of the fraction converge to zero. As sin(pi/2) = 1, it cannot be invoked. So, in short:

1/infinity = 0, and 1/(-infinity) = 0
therefore 1/tan(pi/2) = cot(pi/2) = 0
but
1/undefined = undefined