# Intuitive understanding of limit cos(x)/x as x->0

1. Jun 6, 2010

### Moogie

Hi

I'm new to calculus and self-teaching so please be kind to me :)

The book I am reading says I should convince myself that the limit of cos(x)/x as x->0 does not exist so i thought i'd check my thinking on here.

by very small i mean close to 0

The lim as x->0+ of cos(x)/x:
As x gets very small cos(x) approaches 1
As x gets very small x becomes a small positive number
So 1/tiny positive number = + infinity

The lim as x->0- of cos(x)/x:
As -x gets very small cos(x) approaches 1
As -x gets very small, x becomes a small positive number
So 1/tiny negative number = - infinity

So the left and right handed limits aren't the same so the limit does not exist.

If anyone was willing to show me how you write this properly it would be much appreciated

thanks

2. Jun 6, 2010

### Office_Shredder

Staff Emeritus
Here's a kind of brute force method:

If $$\frac{-1}{2}<x<\frac{1}{n}$$ for n>2, then $$cos(x)<\frac{1}{2}$$ (since $$cos(\frac{\pi}{6}=\frac{1}{2}$$. All you really are interested in here is that cos(x) is bounded away from zero).

Then if $$0<x<\frac{1}{n}$$ we have $$\frac{cos(x)}{x} \geq \frac{n}{2}$$ and if $$\frac{-1}{n}<x<0$$, we have $$\frac{cos(x)}{x} \leq \frac{-n}{2}$$

And then follow up with some conclusions

3. Jun 6, 2010

### vela

Staff Emeritus
If you look at the formal definition of the limit of a function, you'll see that for the limit to exist, (cos x)/x has to approach some real number L as x approaches 0. The fact that (cos x)/x is unbounded as you approach 0 from either side is enough to say that the limit doesn't exist. So when you write something like

$$\lim_{x \to c} f(x) = \infty$$

you're actually saying the limit doesn't exist and you're saying why, i.e. f(x) blows up there.

In comparison, if you look at the function f(x)=|x|/x as x approaches 0, then the one-sided limits do exist, so there's still the possibility that the two-sided limit exists. But in this case, the one-sided limits

$$\lim_{x \to 0^+} \frac{|x|}{x} = 1$$

and

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

aren't equal, so you would conclude from that that the limit doesn't exist.

Last edited: Jun 6, 2010
4. Jun 6, 2010

### Office_Shredder

Staff Emeritus
In this case you can't even say the limit is infinity, because it goes to negative infinity from the other side.

5. Jun 6, 2010

### Moogie

"The fact that (cos x)/x is unbounded as you approach 0 from either side is enough to say that the limit doesn't exist."

Sorry if I am too basic for everyone here but I am teaching myself maths and don't know all of the terms you use. What do you mean by cos x /x is unbounded

6. Jun 6, 2010

### vela

Staff Emeritus
In plain English, it just means (cos x)/x goes to infinity as x goes to 0+. Similarly, the function is also unbounded coming from the left because it goes to -infinity.

To be a bit more technical, if you approach 0 from the right, for example, there's no number M such that (cos x)/x is always less than M. If there were such a number, you'd call M an upper bound, and (cos x)/x would be bounded (from above).

7. Jun 6, 2010

### Moogie

Hi

Thanks for that. Does unbounded specifically mean as you approach 0 from above (0+) because you didn't mention 0-

Was everything correct/ok in the assumptions I made in my original question? I appreciate I could have expressed them in a more formal way but I'm getting used to the language/notation. I'm more concerned about my intuition of the problem

thanks

8. Jun 6, 2010

### vela

Staff Emeritus
No, the function isn't bounded from below when $x \to 0^-$, so it's unbounded on that side too.