# Homework Help: Limits question!

1. Sep 15, 2008

### ashleyrc

1. The problem statement, all variables and given/known data
limit as x approaches 0 of (sin^2 (x))/x^2

2. Relevant 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?

3. The attempt at a solution

2. Sep 15, 2008

### danago

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

3. Sep 15, 2008

### 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.

4. Sep 15, 2008

### Dick

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.

5. Sep 15, 2008

### 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

6. Sep 15, 2008

### danago

Woa hold up. You cant 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.

7. Sep 15, 2008

### 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.

8. Sep 15, 2008

### 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

9. Sep 16, 2008

### 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

10. Sep 16, 2008

### bdforbes

Provided the latter two limits exist

11. Sep 16, 2008

### HallsofIvy

which, fortunately, they do in this case.

12. Sep 16, 2008

### bdforbes

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

13. Sep 17, 2008

### JG89

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.

14. Sep 17, 2008

### bdforbes

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

15. Sep 17, 2008

### 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.

16. Sep 17, 2008

### boombaby

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.

17. Sep 17, 2008

### Dick

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.

18. Sep 18, 2008

### bdforbes

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.

19. Sep 18, 2008

### 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 plz, 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)

Last edited: Sep 18, 2008
20. Sep 18, 2008

### 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.

21. Sep 18, 2008

### 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".

22. Sep 19, 2008

### boombaby

This is pretty cool! Thanks!