# Exceedingly easy limit

1. Aug 5, 2010

### Char. Limit

1. The problem statement, all variables and given/known data
So, I was reviewing limits for some reason, and I came across one that couldn't be solved:

$$\stackrel{lim}{x\rightarrow0} \frac{cos(x)-1}{x^2}$$

2. Relevant equations

Are there any? I'm not sure.

3. The attempt at a solution

Now, I could do this limit easily if I had the Taylor series for cosine at my disposal (using that method, I got -1/2), but I'm trying to do this limit without using the derivative of the cosine, which essentially means Taylor series is out. And considering that limits were sort of just glanced over in my Calc I class, not really delved into, how would I solve this without assuming a derivative for the cosine?

2. Aug 5, 2010

### Dick

If you are living in a world without derivatives, then you'll have to try and use a geometric argument. Draw a unit circle and a right triangle with central angle x, one leg along the x-axis. Use that when x is small the hypotenuse is approximately equal to the arc length. I.e. the hypotenuse is ~x. Do you see the picture? Now use Pythagoras.

3. Aug 5, 2010

### Char. Limit

I'm afraid I don't see it... you say the central angle is x, but then later you say that the hypotenuse is ~x because it's close to the arc length (of which arc?). Which is it, or is it both?

4. Aug 5, 2010

### ehild

You can use the identity 1-cos(x)=2sin2(x/2), and then using the limit (sinx/x) =1 when x-->0

ehild

Last edited: Aug 5, 2010
5. Aug 5, 2010

### Dick

Ack. I should learn to make pictures sometime. The central angle is x (in RADIANS). So the length of the arc subtended by x on a unit circle is x. Since arclength=r*angle. And r is 1. That's very closely equal the chord of the subtended arc with central angle x when x is small. Do you see that? If you wait for me to draw a picture, it might take a few days.

6. Aug 5, 2010

### Dick

Of course Char. Limit does. I think the ground rules here are to do this without using derivatives. If you can help to explain the geometric argument that would be great.

7. Aug 5, 2010

### Char. Limit

Oh, the chord! I thought you meant the arc length approached the hypotenuse length.

8. Aug 5, 2010

### ehild

Rewrite the limit in terms of sin(x/2)/(x/2)and use that it is 1 when x tends to 0. This limit is obtained from the figure attached:
sin(x)<x<tan(x) (x is the arc which can be approximated with the cord) and dividing the inequality by sin(x)

ehild

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9. Aug 5, 2010

### Char. Limit

Well, so far three methods (Taylor Series, L'Hopital's rule, and Ehild's Substitution), and they all gave me -1/2, but for some reason the geometric version gave me -1. I drew the unit circle, a triangle with central angle x, and noticed that as x-->0, the chord was ~x, and the adjacent side was ~1. This made the hypotenuse ~1+x^2, and the cosine of x ~1/(1+x^2). So I substituted that in for cos(x).

$$\frac{\frac{1}{1+x^2} - 1}{x^2}$$

$$\frac{\frac{1}{1+x^2}-\frac{1+x^2}{1+x^2}}{x^2}$$

$$\frac{\frac{1-1-x^2}{1+x^2}}{x^2}$$

$$-\frac{\frac{x^2}{1+x^2}}{x^2}$$

$$-\frac{x^2}{x^2(1+x^2)}$$
$$-\frac{1}{1+x^2}$$

Which tends to -1 as x tends to 0.

10. Aug 5, 2010

### Dick

Actually, that is what I meant. The chord and the arc are approximately equal. So if you've drawn the right triangle you should be looking at sin(x)^2+(1-cos(x))^2~x^2. I'm not sure how rigorous that is but that's how I seem to remember the argument.

11. Aug 5, 2010

### ehild

I can not follow you without picture.
See my one. Calculate the length of chord from the yellow triangle.

ehild

#### Attached Files:

• ###### limit2.JPG
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12. Aug 5, 2010

### Dick

Thank you for the picture, ehild. That's just what I needed. And the sin(x)<x<tan(x) is probably what you need to actually make it rigorous.

13. Aug 5, 2010

### Char. Limit

Wow, I completely misunderstood Dick's description. So, with this new picture, I get that...

$$sin^2(x)+\left(1-cos(x)\right)^2\approx x^2$$

$$sin^2(x) + 1 - 2 cos(x) + cos^2(x) \approx x^2$$

$$2-2cos(x) = 2(1-cos(x)) \approx x^2$$

$$1-cos(x) \approx \frac{x^2}{2}$$

For very small x. Correct? So, putting that in...

$$lim_{x\rightarrow0}{x\frac{cos(x)-1}{x^2} = lim_{x\rightarrow0}-\frac{1-cos(x)}{x^2}$$

$$=lim_{x\rightarrow0}-\frac{\frac{x^2}{2}}{x^2}$$

$$=lim_{x\rightarrow0}-\frac{x^2}{2x^2}=lim_{x\rightarrow0}-\frac{1}{2}=-\frac{1}{2}$$

Thus concludes the proof?

14. Aug 5, 2010

### Bohrok

Wouldn't it be easier to multiply by (cosx + 1)/(cosx + 1) and use the limit for sinx/x?

15. Aug 5, 2010

### ehild

Char, it is more or less OK.

So you start out from chord/arc-->1 for x--->0
You know that arc=x. Determine the length of chord from the triangle:

chord2=2(1-cosx),

2(1-cosx)/x2 --->1

-(1-cosx)/x2-->-1/2

ehild