# Limit as x->0

1. Mar 31, 2008

### caesius

1. The problem statement, all variables and given/known data
Find the limit: lim x->0 (3/x^2 - 2/x^4)

2. Relevant equations

3. The attempt at a solution
I've tried manipulating it in a lot of ways, but nomatter what I do I've still either got the limit of the denominator = 0 or one of the terms in either the num. or den. being divided by x (which is tending to zero)

I've got as far as lim x->0 ( 9x^4 - 4 / 3x^6 + 2x^4 ) but this doesn't help as the den. = 0.

I've graphed this function and had a look - the limit should equal -(infinity).

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

2. Relevant equations

3. The attempt at a solution

2. Mar 31, 2008

### CompuChip

Welcome to PF caesius.

Graphing the function was a good idea, at least now you know the answer

You have
$$\lim_{x\to0} \left( \frac{3}{x^2} - \frac{2}{x^4} \right)$$
Standard approach in these cases: write it as a single fraction. Though in principle you did this correctly, you could have done it much more easily by just multiplying the first term only by $x^2 / x^2$ and then adding the fractions. Of what you then get it is much easier to show that it tends to -infinity.

3. Mar 31, 2008

### caesius

Ok, I get (3x^2 - 2) / x^4 but I don't see how this helps.

If x was tending to infinity I see how to solve it but for 0?

4. Mar 31, 2008

### CompuChip

Divide numerator and denominator by $x^4$.

5. Mar 31, 2008

### caesius

I've done that but I get 3x^-2 - 2x^-2 / 1

Which to me looks like: (undefined) - (undefined) / 1

How can that have a solution?

6. Mar 31, 2008

### HallsofIvy

Staff Emeritus
Isn't "the limit does not exist" a solution?

7. Mar 31, 2008

### caesius

Ok I've just read how to format properly, for clarity I have:

$$\frac{\frac{3}{x^2} - \frac{2}{x^2}}{1}$$

which when taking the limit as x tends to zero will result in indefined operations no?

8. Mar 31, 2008

### caesius

But I know the "limit" is $$-\infty$$, and I'm aware that this is not a limit by definition, surely I should be able to show this from the equation.

9. Mar 31, 2008

### HallsofIvy

Staff Emeritus
Okay, here's what I would have done: combine the two fractions to get
$$\frac{3}{x^2}- \frac{2}{x^4}= \frac{4x^2- 2}{x^4}$$
Since the numerator goes to -2 while the denominator goes to 0, the fraction itself goes to $-\infty$ (i.e. the limit does not exist). What more do you want?