# Thread: Why is (sin x)/x at x=0 equal 1?

1. ## Why is (sin x)/x at x=0 equal 1?

Hi I am doing a simpson's rule problem, and one of the values I need is (sin x)/x at x=o. Now my calculator gives undefined which should be the answer, but graphing on wolfram alpha or on my calculator shows that at x=0 the answer is 1.

2. Originally Posted by anees7112
Hi I am doing a simpson's rule problem, and one of the values I need is (sin x)/x at x=o. Now my calculator gives undefined which should be the answer, but graphing on wolfram alpha or on my calculator shows that at x=0 the answer is 1.
The value of f(x) at x = ξ where ξ is constant is just a coordinate. You can't graph that. Probably the calculator and wolfram are considering the values of f(x) = sinx/x as x goes to 0.

3. Originally Posted by TheCoffeeMachine
The value of f(x) at x = ξ where ξ constant is just a point. You can't graph that. Probably the calculator and wolfram are considering the values of f(x) = sinx/x as x goes to 0.
Alright I get what you mean there, but for the problem I still need a y0 value (Y where x=0). So what should I put as this y0 value?
Thank you

4. Originally Posted by anees7112
Alright I get what you mean there, but for the problem I still need a y0 value (Y where x=0). So what should I put as this y0 value?
Thank you
You won't be able to put a finite value at $\displaystyle \displaystyle x = 0$, because as has been stated, even though $\displaystyle \displaystyle \lim_{x \to 0}\frac{\sin{x}}{x} = 1$, the function is undefined at $\displaystyle \displaystyle x = 0$.

If you MUST input a value at $\displaystyle \displaystyle x = 0$, you will need to change your function to

$\displaystyle \displaystyle \begin{cases} \frac{\sin{x}}{x} &\phantom{|}\textrm{if }x \neq 0\\ 1 &\phantom{|}\textrm{if }x = 0 \end{cases}$

5. In the XVIII° century the Swiss mathematician Leonhard Euler demonstrated that the function $\displaystyle \frac{\sin x}{x}$ is well represented as 'infinite sum' or 'infinite product'...

$\displaystyle \frac{\sin x}{x}= 1+ \sum_{n=1}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} = \prod_{n=1}^{\infty} (1-\frac{x^{2}}{n^{2}\ \pi^{2}})$ (1)

... and as consequence he obtained the important result...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}= \frac{\pi^{2}}{6}$ (2)

Now if we set x=1 in the 'infinite sum' or in the 'infinite product' we obtain in both cases 1... so that, in my opinion [and probably in Euler's opinion...] , the statement that the function $\displaystyle \frac{\sin x}{x}$ in x=0 is 'indeterminate' is highly questionable...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. well i guess it depends on what you mean by "equality".

consider the function $\displaystyle f(x) = \frac{x^2-1}{x-1}$. what is f(1)? is it "determinate"?

a function with a removable discontinuity is not quite the same thing as a continuous function.

i believe it is important to distinguish between a limit, and its value.

7. Originally Posted by Deveno
well i guess it depends on what you mean by "equality".

consider the function $\displaystyle f(x) = \frac{x^2-1}{x-1}$. what is f(1)? is it "determinate"?

a function with a removable discontinuity is not quite the same thing as a continuous function.

i believe it is important to distinguish between a limit, and its value.
I would usually define the equality of functions as the following: two functions are equal if their domains are the same, and their rules of association are the same.

8. well you often hear of people talking about "the function x^2" (or in this case, sin(x)/x). "x^2" isn't a function, it's a number. a better "name" for the function would be [ ]^2, or something similar, but such a notation gets cumbersome. if all you are given is the "rule" for the function, the standard practice is to allow the domain to be "everything within the realm of discourse that makes sense". for example, in calculus, this is "all real numbers for which f(x) is defined".

sin(x)/x is undefined at 0. so its domain is R-{0}. it has no "value" at 0. it does have a limit at 0, and the function resulting from using the value of the limit at x = 0 is even continuous. but as functions the two are not quite equal (different domains).

but the problem with chisigma's approach is even more subtle. his equation has a number on the left-hand side, and an infinite sum on the right hand side. the right-hand side is actually a limit in disguise (the limit of the partial sums, or in the second expression, the partial products). now, even though the radius of convergence is all of R, there is a slight wrinkle here: are we using the "equals" sign to indicate numerical equality, or as a definition? because as numerical equality, sin(x)/x is still undefined at 0, 0/0 is not a real number. as a definition (for example, if one uses Taylor series to actually define the sin function, which is one possible approach), there is no problem (the x's "cancel out").

this subtle but important distinction is underscored by the second sum involving pi. notice that the terms in the infinte sum are all rational, while the "sum" is irrational. the set of limit points of the set of rational numbers is "bigger" than the set of rational numbers. what this meant was not well-understood by the founders of calculus, possibly not even by euler himself. these type of considerations are, at their heart, no longer algebraic, but topological, which is what gives calculus its particular and singular flavor.

,

,

,

,

### is (sin0)/0 defined????

Click on a term to search for related topics.