Limiting Behavior of sin^2(x)/x^2 at x=0

[ashleyrc]

Homework Statement

limit as x approaches 0 of (sin^2 (x))/x^2

Homework Equations

i generally know how to solve the equation, but I'm not sure what to do about the top. is sin^2 (x) the same as sinx^2?

The Attempt at a Solution

 

[danago]

Homework Statement

limit as x approaches 0 of (sin^2 (x))/x^2

Homework Equations

i generally know how to solve the equation, but I'm not sure what to do about the top. is sin^2 (x) the same as sinx^2?

The Attempt at a Solution

sin^2 (x) = (sin x)^2

 

[bdforbes]

You have probably learned a rule for the situation where both the numerator and denominator go to zero (or both to infinity). That is what I would use.

 

[Dick]

Homework Statement

limit as x approaches 0 of (sin^2 (x))/x^2

Homework Equations

i generally know how to solve the equation, but I'm not sure what to do about the top. is sin^2 (x) the same as sinx^2?

The Attempt at a Solution

sinx^2 can mean two different things. It could mean sin(x^2) or (sin(x))^2. Which do you mean? Try not to write ambiguous forms when posting a question. As danago said, sin^2(x) is the latter. And bdforbes is referring to l'Hopital's rule. If you haven't learned that then you probably proved directly that sin(x)/x approaches 1 as x approaches 0. Use that.

 

[ashleyrc]

ok, so it would = (sin x)^2/ x^2, then the limit is 1. because you can cancel the ^2 on top and bottom to = (sin x)/x, which is 1

 

[danago]

ok, so it would = (sin x)^2/ x^2, then the limit is 1. because you can cancel the ^2 on top and bottom to = (sin x)/x, which is 1

Woa hold up. You can't just cancel the indexes from the top and the bottom. Id probably solve this by using the result that Dick mentioned above with one of the general limit rules.

 

[Dick]

2^2/1^2=4. 2/1=2. You CAN'T cancel an exponent. You can write it as (sin(x)/x)^2 and draw conclusions from that.

 

[ashleyrc]

yeah, i remember the preofesor mentioning l hopitals rule, but i don't really know it yet, i'll look it up and get back w/u all later

 

[JG89]

Use these two pieces of information:

1) (sinx)^2/x^2 = (sinx/x) * (sinx/x)

2) Given two functions, f(x) and g(x, and any point p, the lim [f(x) * g(x)] = lim f(x) * lim g(x)
x-> p x-> p x-> p

 

[bdforbes]

Use these two pieces of information:

1) (sinx)^2/x^2 = (sinx/x) * (sinx/x)

2) Given two functions, f(x) and g(x, and any point p, the lim [f(x) * g(x)] = lim f(x) * lim g(x)
x-> p x-> p x-> p

Provided the latter two limits exist

 

[HallsofIvy]

which, fortunately, they do in this case.

 

[bdforbes]

Now the OP has only to Taylor expand sin(x) and the problem will be solved.

 

[JG89]

Now the OP has only to Taylor expand sin(x) and the problem will be solved.

I don't see a need for that. It seems to me that it's sufficient to rewrite (sinx)^2/x^2 as (sinx/x) * (sinx/x) and to remember that the limit of a product of functions is the product of each limit.

 

[bdforbes]

I don't see a need for that. It seems to me that it's sufficient to rewrite (sinx)^2/x^2 as (sinx/x) * (sinx/x) and to remember that the limit of a product of functions is the product of each limit.

Yes but what is the limit of sinx/x? I propose a Taylor expansion or l'Hopitals rule.

 

[JG89]

0 < cosx < (sinx)/x < 1/cosx is valid for -pi/2 < x < pi/2. Using the squeeze theorem here is extremely quick to find the limit of sinx/x as x approaches 0.

 

[boombaby]

Yes but what is the limit of sinx/x? I propose a Taylor expansion or l'Hopitals rule.

lim(sinx)/x=1 should exists before we get (sinx)'=cosx. Thus L'hospital's rule is not a proper way to prove(though it's a way to calculate) that sinx/x->1, as x->0.

0 < cosx < (sinx)/x < 1/cosx is valid for -pi/2 < x < pi/2. Using the squeeze theorem here is extremely quick to find the limit of sinx/x as x approaches 0.

It works.

 

[Dick]

lim(sinx)/x=1 should exists before we get (sinx)'=cosx. Thus L'hospital's rule is not a proper way to prove(though it's a way to calculate) that sinx/x->1, as x->0.It works.

Gack! You don't have to know the limit exists to apply l'Hopital. All you have to know is that it has the form 0/0. Why I'm contributing again to this bloated useless thread, I don't know. But that is soooo wrong.

 

[bdforbes]

lim(sinx)/x=1 should exists before we get (sinx)'=cosx. Thus L'hospital's rule is not a proper way to prove(though it's a way to calculate) that sinx/x->1, as x->0.

I disagree. If $$\stackrel{lim}{x\rightarrow0}\frac{f'(x)}{g'(x)}$$ exists, and $$\stackrel{lim}{x\rightarrow0}f(x)=0$$ and $$\stackrel{lim}{x\rightarrow0}g(x)=0$$, then it is true that $$\stackrel{lim}{x\rightarrow0}\frac{f(x)}{g(x)} = \stackrel{lim}{x\rightarrow0}\frac{f'(x)}{g'(x)}$$. Do you not agree with this? The problem at hand satisfies these conditions, so I don't see why we couldn't use it.

 

[boombaby]

Sorry if I misunderstood someone's words (English is not my mother tongue) . It seems to me that someone is PROVING that lim sinx/x = 1 using L'hospital's rule, which I think is wrong.

If I do not know lim sinx/x = 1 already, how do I know (sinx)' EQUALS TO cosx ( by what means I can get this? If you have one without using "lim sinx/x = 1", tell me please, just curious about it )? If I do not know that (sinx)' equals to cosx, how do I apply L'hospital's rule?

As I said before, L'hospitals is okay to CALCULATE sinx/x but is not a PROPER way to PROVE it. They are absolutely different things.

Applying the L'hospital's rule to the original question is nothing wrong, since we simply take "lim sinx/x = 1" as true, and moreover, "(sinx)'=cosx" as true.

Anything wrong? Sorry again if I misunderstood something

just to make it more clear. Try to think that we are asked to prove that (sinx)'=cosx. We want to know what is the result of lim (sin(x+h)-sinx )/h when h->0, and you are not going to use L'hospital's rule to prove it, right? (If using L'hospital's rule, we of course get the correct answer cosx! but unfortunately, we are using "(sinx)'=cosx" already)

 

[bdforbes]

Actually what we want to know is lim sinx/x. We are allowed to assume (sinx)'=cosx, this isn't high school maths. Then l'Hopitals rule DOES prove the limit.

 

[HallsofIvy]

There are a number of different ways of DEFINING sin(x) and cos(x) so saying that you must prove lim sin(x)/x= 1 before you can show that sin'(x)= cos(x) is not true.

For example, my favorite way of defining sin(x) and cos(x) is:

sin(x) is the function satisfying the initial value problem y"= -y with y(0)= 0, y'(0)= 1.

cos(x) is the function satisfying the initial value problem y"= -y with y(0)= 1, y'(0)= 0.

We know from the existence and uniqueness theorem for initial value problems that there exist such functions and that they are unique. It follows from the fact that they satisfy a second order differential equation that sin(x) and cos(x) are at least twice differentiable.

Furthermore, since that is a linear, homogeneous, differential equation we know that ANY solution to the equation can be written as a linear combination of two independent solutions. That sin(x) and cos(x) are independent is easy to show: if Asin(x)+ Bcos(x)= 0 for all x, then, in particular, if x= 0, A sin(0)+ B cos(0)= B= 0. With B= 0, we have Asin(x)= 0 for all x. But sin(x) is NOT 0 for all x because its derivative at 0 is not 0. Thus A= 0 and sin(x) and cos(x) are independent and any solution to the equation can be written in the form A sin(x)+ B cos(x).

In fact, suppose y(x) is a solution to the equation y"= -y. Then y(x)= A sin(x)+ B cos(x) for some A and B. Then y is differentiable and y'(x)= A sin'(x)+ B cos'(x). Since sin(0)= 0 and cos(0)= 1, y(0)= A. Since sin'(0)= 1 and cos'(0)= 0, y'(0)= B. That is, the coefficients, A and B, are precisely y(0) and y'(0).

Now, let y(x)= sin'(x). Then, since sin(x) is twice differentiable, y is differentiable and y'(x)= sin"(x). But sin(x) satisfies y"= -y so sin"(x)= -sin(x): y'(x)= - sin(x). Since sin(x) is differentiable, y is twice differentiable and we have y"(x)= - sin'(x)= -y. That is, y(x) satisfies the same differential equation, y"= -y. y(0)= sin'(0)= 1 and y'(0)= -sin(0)= 0. That is sin'(x)= y(x)= y(0)cos(x)+ y'(0)sin(x)= cos(x).

Let y(x)= cos'(x). Again, we can show that y'(x)= cos"(x)= - cos(x) and that y satisfies y"= -y so y(x)= y(0)cos(x)+ y'(0)sin(x) and y(0)= cos'(0)= 0 and y'(0)= - cos(0)= -1. That is cos'(x)= y(x)= (0)cos(x)+ (-1) sin(x)= -sin(x).

So it is NOT necessary to have first learned "lim sin(x)/x= 1" in order to get the derivatives of sin(x) and cos(x). I do agree, however, that, for a person who has learned the standard "circular functions" definition of sin(x) and cos(x), L'Hopital's rule is "overkill".

 

[boombaby]

This is pretty cool! Thanks!