# Homework Help: Help with a limit?

1. Jun 4, 2004

I’m having trouble understanding this limit

Lim x->0 (x^2-2|x|)/x

I think its undefined but lost marks for that answer on my class assignment.

Also, Why can’t I simplify to x-2?

2. Jun 4, 2004

### AKG

The limit exists if and only if the left limit and right limit exist, and are equal to each other.

$$\lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2|x|}{x}$$

$$= \lim _{x \rightarrow {0}^{+}} \frac{x^2 - 2x}{x}$$

$$= \lim _{x \rightarrow {0}^{+}} x - 2$$

$$= -2$$

$$\lim _{x \rightarrow {0}^{-}} \frac{x^2 - 2|x|}{x}$$

$$= \lim _{x \rightarrow {0}^{-}} \frac{x^2 + 2x}{x}$$

$$= \lim _{x \rightarrow {0}^{-}} x + 2$$

$$= 2$$

Therefore, the limit does not exist.

Last edited: Jun 4, 2004
3. Jun 4, 2004

### TALewis

You can also see it graphically by plotting the function. There is a large discontinuity (gap) at x=0.

4. Jun 4, 2004

### e(ho0n3

Say x = -x. Then
$$\frac{(-x)^2 - 2|-x|}{-x} = \frac{x^2 - 2x}{-x} = 2 - x$$​
As you can see, you can't JUST simplify to x - 2. If x > 0, then you can simplify to x - 2.

What happens when x approaches 0 from the positive direction? What happens when x approaches 0 from the negative direction? You should see (if you haven't already) that the limit is undefined.

5. Jun 4, 2004

### Parth Dave

because |x|/x is not equal to 1. It is equal to 1 if x >0 and -1 if x < 0. You can simplify that to x - 2|x|/x. That would help in finding the limit. In order to find the limit you will have to break that up into a piecewise function and than look at the limit as it approaches 0 from the left and the right. And ill give you a hint, there is no limit as x --> 0.