Expert: Paul Klarreich - 2/9/2011

Question
QUESTION: When x = 0 is it true that sin(x)/x = 1 or is this just the case as x tends to zero. I feel the curve sin(x)/x is not continuous, but there is an infinitely small gap at x = 0.

My reason for this is that Taylor is just an approximation and not absolute.

ANSWER: Questioner: Trevor
Country: United Kingdom
Category: Calculus
Private: No
Subject: Is sinx/x a continous curve
Question: When x = 0 is it true that sin(x)/x = 1 or is this just the case as x tends to zero. I feel the curve sin(x)/x is not continuous, but there is an infinitely small gap at x = 0.

My reason for this is that Taylor is just an approximation and not absolute.
..............................................
sin 0
-----  is undefined.
 0

         sin x
lim[x->0] ----- = 1,
         x

as you will find in any text.

When you apply the definition of continuity, you will see that it fails the first test,  i.e.  f(0) is not defined.

So you will conclude that it is not continuous at x = 0.

AND you will try to avoid non-mathematical language like:

"I feel the curve"

and

"infinitely small gap"

"Taylor is just an approximation" <Taylor is (or was) a person>

Learn and use the vocabulary and you will "feel much better" about mathematics.


---------- FOLLOW-UP ----------

QUESTION: Are you saying that this is not a continuous graph at x = 0.  
What is the test for continuity that this fails to meet?

When I referred to Taylor I meant the "Taylor series" that shows that when x = 0, the graph is equal to 1.

What is the correct terminology to use, so that I can use the the correct mathematical language when writing up my findings.

Answer
QUESTION: Are you saying that this is not a continuous graph at x = 0.  

Yes.

What is the test for continuity that this fails to meet?

I wrote that in the answer:

"When you apply the definition of continuity, you will see that it fails the first test,  i.e.  f(0) is not defined."

Look up the definition in your text -- it's there.

................................

There is a 'Taylor approximation' which is a polynomial with remainder and:

There is a 'Taylor series' which is a representation of the function.

Now the T.S. for sin x is :

x - x^3/3! + ... + (-1)^n x^(2n+1)/(2n+1)! + ...

Now you want to divide that by x, I suppose, to get:

sin x
----- = (so we say)
 x

1 - x^2/3! + ... + (-1)^n x^(2n)/(2n+1)! + ...

and then you want to put  x = 0 and say all the terms become zero except the first one, right?

So you will say the answer is sin x / x = 1

I leave it to you to find the defect in that argument.