# Homework Help: Limit problem

1. Jan 5, 2012

### lab-rat

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

Evaluate limx->0 (e^x - 1- x - (x^2/2))/x^3

3. The attempt at a solution

I can't remember how to solve this limit. Do I need to evaluate each part seperately? I plugged in the 0 to find that the limit does exist. I just can't seem to figure out what to do next.

2. Jan 5, 2012

### BloodyFrozen

Why not use l'Hôpital's rule?

3. Jan 5, 2012

### micromass

Or use the Taylor series of the exponential function.

4. Jan 5, 2012

### lab-rat

If I use l'Hopital's rule won't I end up with 3x^2 at the bottom?

5. Jan 5, 2012

### micromass

You will want to use l'Hopital multiple times.

6. Jan 6, 2012

### genericusrnme

The limit of a sum is the sum of limits - Lim[a+b]=Lim[a]+Lim
Then as everyone before has said, you're going to want to use l'hopitals rule for the last term

7. Jan 7, 2012

### lab-rat

I used L'Hopital's rule three times and ended up with limx->0(e^x)/5

Is this correct? It still gives me 0 on top.

8. Jan 7, 2012

### lab-rat

Here is how I did it in case it might help

lim->0 (e^x)' - 1' - x' - (x^2/2)' /x^3'

= lim->0 (e^x' - 1' - x')/ 3x^2'

= lim->0 (e^x' - 1') / 5x'

= lim x->0 (e^x) /5

9. Jan 7, 2012

### BloodyFrozen

Very close, but look at your second to thrid line in the denominator.

What is: $$\frac{d}{dx}(3x^{2})$$

After you fix that, evaluate it at 0.

$$\lim_{x\to 0}~ e^{x}$$ should not be 0.

Last edited: Jan 7, 2012
10. Jan 7, 2012

### lab-rat

Thanks!
It should be 6x right?

So is 1/6 the correct answer?

11. Jan 7, 2012

### SammyS

Staff Emeritus
1/6 is the correct result.