# Homework Help: Calculating a limit with tangent as denominator

1. Feb 16, 2012

### Tebow15

1. The problem statement, all variables and given/known data
lim x^3-2x^2+x/tanx
x->0

3. The attempt at a solution

All i know is that tan is going to break up into sinx/cosx so the equation will look like this

lim x^3-2x^2+x/(sinx/cosx)
x->0

I haven't worked with cubic or quadratic functions yet so I don't know where to begin with this.

2. Feb 16, 2012

### Mentallic

Remember to throw in brackets where necessary, because the way it looks right now is

$$\lim_{x\to 0}\left(x^3-2x^2+\frac{x}{\tan(x)}\right)$$

So you need to put brackets around the cubic to avoid this confusion

If your problem is of the form $$\lim_{x\to a}\left(f(x)\cdot g(x)\right)$$ then this is equivalent to $$\lim_{x\to a}f(x)\cdot\lim_{x\to a} g(x)$$

So try factoring out an x from the numerator, and see if you can apply this idea correctly.

3. Feb 16, 2012

### Tebow15

I may have confused you with the way I wrote it, it should look like this. (I'll use brackets to make it clearer)

lim (x^3-2x^2+x)/(sinx/cosx)
x->0

4. Feb 16, 2012

### Tebow15

The answer eventually = 1 but I don't know how to get there

5. Feb 17, 2012

### ehild

Factor out x as Mentallic suggested and rearange the expression as

$$\left(\frac{x}{\sin(x)}\right)(x^2-2x+1)\cos(x)$$

ehild