# Limx->0 (sinx/x)=1?

1. Jun 11, 2012

### charmedbeauty

limx-->0 (sinx/x)=1?

1. The problem statement, all variables and given/known data

Hi I have the question evaluate the limit going to zero of

(cos(x)-1)/sin(x2)

2. Relevant equations

3. The attempt at a solution

lim x→0 (cos(x)-1)/sin(x2)

goes to 0/0 by Lhoptal's

f'(x)/g'(x)

= (limx→0 sinx/x)(limx→0 -1/ 2cos(x2))

but the answer I have says that (limx→0 sinx/x) = 1

how is this so

I know the other term goes to -1/2 but yeah I'm really confused about the sin term going to 1 is this a mistake or is there something i don't know?

thanks.

2. Jun 11, 2012

### Dickfore

Re: limx-->0 (sinx/x)=1?

Hint:
$$\cos x - 1 = -2 \sin^2 \left( \frac{x}{2} \right)$$
Divide the numerator and denominator by $x^2$, and use your limit.

3. Jun 11, 2012

### charmedbeauty

Re: limx-->0 (sinx/x)=1?

do you mean cos2x-1=-2sin2(x/2)?

if i divide numerator and denominator by x2

(sinx × x2)/x3

I still get 0 out of it when I plug 0 into anything times by sin i get 0.

do I take L'hopitals again?

of the individual sin part of the function? can you partially diff and apply L'hopitals?

4. Jun 11, 2012

### Dickfore

Re: limx-->0 (sinx/x)=1?

No, because:
a) it is an incorrect trigonometric identity;
b) it is not useful for your problem, since the left-hand side is not what you have in your limit.

I meant exactly what I wrote:
$$\cos x - 1 = -2 \, \sin^2 \left( \frac{x}{2} \right)$$

I don't know how you got this.

5. Jun 11, 2012

### clamtrox

Re: limx-->0 (sinx/x)=1?

What happens if you try L'Hopital again?

6. Jun 11, 2012

### HallsofIvy

Staff Emeritus
Re: limx-->0 (sinx/x)=1?

I think I would be inclined to write it as
$$\frac{cos(x)- 1}{sin(x^2)}= \frac{x^2}{sin(x^2)}\frac{cos(x)- 1}{x^2}$$

It's easy to see that $sin(x^2)/x^2$ goes to 1. Using L'Hopital's rule on $$\frac{cos(x)- 1}{x^2}$$ leads to $$\frac{-sin(x)}{2x}$$ which has limit -2.

7. Jun 11, 2012

### charmedbeauty

Re: limx-->0 (sinx/x)=1?

Ok, I can see how you get that but the part I'm confused with is that you have a cos factor in the numerator and a sin factor in the denominator. every time L'hopital's gets used these just swap from sin to cos visca versa.

and cos 0 =1 but sin0=0

thats how I don't understand how the zero term vanishes?

can you use L'hopitals rule on just part of a fraction?

because it looks like every time I differentiate tops and bottoms I have a sin and a cos factor, when i really only can have cos factors.?

8. Jun 11, 2012

### Muphrid

Re: limx-->0 (sinx/x)=1?

This is much more easily solved with Taylor expansion.

$$\cos x - 1 = \left(1 - \frac{x^2}{2} + \ldots{}\right) - 1 = -\frac{x^2}{2} + \ldots{}$$

And for the denominator,

$$\sin x^2 = x^2 - \frac{x^6}{6} + \ldots{}$$

So the limit can be written as

$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x\to 0} \frac{-x^2/2 + \ldots}{x^2 - x^6/6 + \ldots{}}$$

The limit is as $x \to 0$, so throw away all but the lowest power term in both the numerator and denominator, and you clearly get

$$\lim_{x \to 0} \frac{-x^2/2}{x^2} = -\frac{1}{2}$$

Now, you originally asked how it could be that,

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

Again, this is easily solved by Taylor expansion.

$$\lim_{x \to 0} \frac{x - x^3/6 + \ldots}{x} = \lim_{x\to 0} \frac{x}{x} = 1$$

Or by l'Hopital's rule,

$$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$$

9. Jun 11, 2012

### HallsofIvy

Staff Emeritus
Re: limx-->0 (sinx/x)=1?

I'm not sure what you mean by that. "vanishes" means "is 0".

You understand what L'Hopital's rule is, dont you?
$$\lim_{x\to a}\frac{f(x)}{g(x)}= \frac{\lim_{x\to a}f'(x)}{\lim_{x\to a} g'(x)}$$
provided those limits exist. I dont know what you mean by "part of a fraction". If you can write a fraction as a sum or product, then you can use $\lim_{x\to a} (f+g)(x)= \lim_{x\to a}f(x)+ \lim_{x\to a}g(x)$ and $\lim_{x\to a}fg(x)= \left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a}g(x)\right)$.

Again, I really don't know what you mean by this. Why would you think "I really only can have cos factors"?

10. Jun 11, 2012

### vela

Staff Emeritus
Re: limx-->0 (sinx/x)=1?

This method likely requires material the OP doesn't yet know considering he or she is still learning how to evaluate basic limits.

11. Jun 11, 2012

### charmedbeauty

Re: limx-->0 (sinx/x)=1?

yep.

12. Jun 11, 2012

### charmedbeauty

Re: limx-->0 (sinx/x)=1?

ok so its ok to have...

limx→0 (-1/2cos(x2)limx→0(cos(x)/1)

= -1/2 (1) =-1/2

??

13. Jun 11, 2012

### vela

Staff Emeritus
Re: limx-->0 (sinx/x)=1?

Applying L'Hopital's rule to what you started with, you got
$$\lim_{x \to 0} \frac{\cos x - 1}{\sin x^2} = \lim_{x \to 0} \frac{-\sin x}{2x\cos x^2}$$ I think what you're missing is this next step. What happened was they broke the single limit into a product of two limits:
$$\lim_{x \to 0} \frac{-\sin x}{2x\cos x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)\left(\lim_{x \to 0} \frac{-1}{2\cos x^2}\right)$$ which you can do if the limits exist. So while it looked like L'Hopital's rule was only being applied to part of the fraction, namely, the sin x/x part, it really wasn't. It's just that the original limit was split into two separate limits. Now the first limit you can evaluate using L'Hopital's rule again.

14. Jun 12, 2012

### charmedbeauty

Re: limx-->0 (sinx/x)=1?

Thanks vela that completely cleared it up.