Question

if you enter

http://www.wolframalpha.com/input/?i=%28sin+x%29+%2F+x+at+x+%3D+0

into wolfram alpha, why does it return 1?

Answer 1

The limit of that expression as x->0 is in fact 1. You can demonstrate this with L'Hopital's rule, as both numerator and denominator approach 0 as x->0.

\[\lim_{x \rightarrow 0} \frac{\sin x}{x} \rightarrow \lim_{x \rightarrow 0} \frac{d(\sin x)/dx}{d(x)/dx} = \lim_{x \rightarrow 0} \frac{\cos x}{1} = 1\]

Answer 2

Right, the limit is 1, but why is the expression evaluated to 1? It should be indeterminate, shouldn't it ?

Answer 3

Shouldn't \[\frac{ \sin 0 }{ 0 }\]

be indeterminate ?

Answer 4

Here's a graph of the sinc function (another name for 1/x * sin(x)):

Answer 5

Maybe one of those analysis geeks can tell you in a convincing fashion why, I'll just tell you it is :-)

Answer 6

Division by 0 is illegal, and so x = 0 is not in the domain of (sin x) / x
Likewise, 0 is not in the domain of the sinc function either.

But my problem is, wolfram keeps telling me the function evaluated at 0 is equal to 1. This is misleading because if you also ask Wolfram to compute the derivative of (sin x) / x at x = 0, it will tell you that it is 1.

(sin x) / x is not continuous at x = 0 and therefore does not have a derivative at x = 0.

Answer 7

There is a series expansion for the function that doesn't involve dividing by 0...

Answer 8

And sinc(0) is defined to equal 1 by at least some authors. http://mathworld.wolfram.com/SincFunction.html

Answer 9

If you've got a quibble with a WA result, send them some feedback with the link at http://www.wolframalpha.com/contact.html

Answer 10

That's a little different, to define the function at a point where it is discontinuous. But that doesn't solve the problem of (sin x) / x. Unless we just define the function to compensate for the discontinuity.

I mean, wolfram does similar calculations for

\[f(x) = \frac{ x^{2} - 9 }{ x - 3 }, f'(3) = 1\]

Yet this function is not defined at x = 3.

Answer 11

i dont have a quibble. i just want to know if i am right to conclude that wolfram gives a wrong response. if i am wrong, just need an explanation why.

Answer 12

Well, I suspect the reasoning is that most people using that expression are in fact interested in the sinc function, and the code treats them interchangeably. Again, the people at Wolfram ought to be in a position to give a definitive answer to such questions about how and why it does what it does :-)