Why is the derivative of sin x / x at x=0 1?

[veneficus5]

Homework Statement

Why is sin x / x at x = 0 equal to 1?

Homework Equations

d/dx (sinc(x)) evaluated at x = 0.

The Attempt at a Solution

El hopital's rule
Cos(x)/1, cos(0) = 1

However when looking at the picture of Sinc, http://www.wolframalpha.com/input/?i=sin(x)/x, it appears that the derivative should be 0 at 0.

Thanks

 

[Dick]

Homework Statement

Why is sin x / x at x = 0 equal to 1?

Homework Equations

d/dx (sinc(x)) evaluated at x = 0.

The Attempt at a Solution

El hopital's rule
Cos(x)/1, cos(0) = 1

However when looking at the picture of Sinc, http://www.wolframalpha.com/input/?i=sin(x)/x, it appears that the derivative should be 0 at 0.

Thanks

If you apply l'hospital's rule to sin(x)/x it doesn't give you the derivative of sin(x)/x at 0. It gives you the limit as x->0 of sin(x)/x.

 

[lurflurf]

write sin(x)/x as either
sin(x)/x=1-x^2/3!+x^4/5!-x^6/7!+...(-1)^k x^(2k)/(2k+1)!+...
or
sin(x)/x=integral t=0 to t=1 cos(x t) dt
$$\int_0^1 \cos(x t) dt$$
and this should be obvious.

 

[veneficus5]

write sin(x)/x as either
sin(x)/x=1-x^2/3!+x^4/5!-x^6/7!+...(-1)^k x^(2k)/(2k+1)!+...
or
sin(x)/x=integral t=0 to t=1 cos(x t) dt
and this should be obvious.

Ah thanks, but I don't understand the last one with the cos.

 

[lurflurf]

^Did you get to integrals and series yet?
You could also write
(sin(x)/x)'=x^-2 (x-tan(x))sec(x)

 

[veneficus5]

write sin(x)/x as either
sin(x)/x=1-x^2/3!+x^4/5!-x^6/7!+...(-1)^k x^(2k)/(2k+1)!+...
or
sin(x)/x=integral t=0 to t=1 cos(x t) dt
$$\int_0^1 \cos(x t) dt$$
and this should be obvious.

Ya I got to them, and I understand the series answer. So I take the derivative of $$\int_0^1 \cos(x t) dt$$ wrt x. This gives me $$\int_0^1 \sin(x t) / -t dt$$. Then what do I do. Sorry it's not obvious :(

 

[Char. Limit]

Ya I got to them, and I understand the series answer. So I take the derivative of $$\int_0^1 \cos(x t) dt$$ wrt x. This gives me $$\int_0^1 \sin(x t) / -t dt$$. Then what do I do. Sorry it's not obvious :(

$$\frac{sin(x)}{x} = \int_0^1 \cos(x t) dt$$

Therefore:

$$\frac{sin(x)}{x}, (x=0) = \int_0^1 \cos(0 t) dt = \int_0^1 dt = 1$$

EDIT: Also, your problem when you're "looking at the sinc function" is that you AREN'T. You're looking at the function sin(x)/x.

 

[veneficus5]

ah I got it. Pretty clever. Thanks

So I can think of el hopital taking the slope of the top and bottom of the fraction. For sin x / x, sin x superimposed upon x makes it clear that it is 1.

It also sort of makes sense how el hopitals only works when there is a singularity on the bottom, but is there a better explanation?

I was messing around with lines like (x+5)*(x-3)/(x-3) and seeing how hopitals worked and didn't work.

I was also a little confused on whether I wanted to derivative or

 

[Mark44]

ah I got it. Pretty clever. Thanks

So I can think of el hopital taking the slope of the top and bottom of the fraction. For sin x / x, sin x superimposed upon x makes it clear that it is 1.

I don't understand what you're saying, "sin x superimposed upon x makes it clear that it is 1."

L'Hopital's Rule applies to limit expressions with the indeterminate forms of $$\left[\frac{0}{0}\right]$$ or $$\left[\frac{\infty}{\infty}\right]$$

BTW, this name is pronounced lo-pi-tal, not el hopital.

If you have a limit expression like this,
$$\lim_{x \to a}\frac{f(x)}{g(x)}$$
where both f(x) and g(x) approach 0 as x approaches a, you can't evaluate the limit, as it is indeterminate.

If, however,
$$\lim_{x \to a}\frac{f '(x)}{g'(x)}$$
has a limit, then that value is the same as the one you really want; namely,
$$\lim_{x \to a}\frac{f(x)}{g(x)}$$

It also sort of makes sense how el hopitals only works when there is a singularity on the bottom, but is there a better explanation?

I was messing around with lines like (x+5)*(x-3)/(x-3) and seeing how hopitals worked and didn't work.

I was also a little confused on whether I wanted to derivative or

Try this one:
$$\lim_{x \to 3}\frac{x^2 - 9}{x - 3}$$

It's probably overkill to use L'Hopital's Rule on this one, as you can also do it by factoring the numerator and cancelling the like terms top and bottom.

 

[veneficus5]

I don't understand what you're saying, "sin x superimposed upon x makes it clear that it is 1."

If you look at the graphs on top of each other, sin x starts out the same as x. So it makes sense that as x approaches 0, one thing divided by the same one thing should be 1.

Thanks for the pronunciation tip. I never knew how to say l'hospital's.

 

[Mark44]

If you look at the graphs on top of each other, sin x starts out the same as x. So it makes sense that as x approaches 0, one thing divided by the same one thing should be 1.

Another way to say this is that for x close to 0, sin(x) $\approx$ x.

Thanks for the pronunciation tip. I never knew how to say l'hospital's.

 

[Rudeboy37]

If you look at the graphs on top of each other, sin x starts out the same as x. So it makes sense that as x approaches 0, one thing divided by the same one thing should be 1.

Be careful about this type of logic. 0/0 (one thing divided by the same one thing) doesn't always equal one. Hence the reason why we have L'Hopital's rule.

Different examples of "0/0" not equaling 1:

lim x->0 (sinx)/(x^3)=infinity

lim x->0 (sinx)/(x^2)=DNE

lim x->0 (x^2)/(sinx)=0

 

[lurflurf]

^Why do people like L'Hopital's rule so much? Saying sin(x)=x+O(x^3) is enough to conclude sin(x)/x->1.

 

[Rudeboy37]

^Why do people like L'Hopital's rule so much? Saying sin(x)=x+O(x^3) is enough to conclude sin(x)/x->1.

Mainly people "like L'Hopital's so much" because many people learn L'Hopital's in Calculus I where as Taylor Series is usually saved for a Calculus II class. Also, many books (like Stewart and Rogawski) cover L'Hopital's at least a couple chapters before Taylor Series are even mentioned. Most students (though not all) trying to evaluate this limit for a class would not know the Taylor Series of sinx, nor what a Taylor Series even means.

 

[Rudeboy37]

However when looking at the picture of Sinc, http://www.wolframalpha.com/input/?i=sin(x)/x, it appears that the derivative should be 0 at 0.

Btw, you originally mentioned the derivative of (sinx)/x at x=0. However, does it even make sense to talk about the derivative at x=0? What must be true for the derivative to even exist at a certain x-value?

 

[HallsofIvy]

Indeed, the title of this thread is an error. Clearly the OP is talking about the limit, not the derivative. Since the function value does not even exist at x= 0, the function is not continuous there and so not differentiable.

How you show that $\lim_{x\to 0} sin(x)/x= 1$ depends upon exactly how you have defined sin(x)! Most calculus texts define sine and cosine in terms of coordinates on the unit circle, then give a rather hand-waving geometric argument. However, you can also define sine and cosine as solutions to certain initial value problems or as power series, then derive other formulas from that.