# Determining a functions this is continuous at 0

1. Jul 3, 2012

### fireychariot

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

f(x) = {2x2 + x +3, x < 0
$\frac{3}{x + 1}$ x ≥ 0

The 2 should be wrapped as 1 with a { but do not know how to do that.

2. Relevant equations

3. The attempt at a solution

I was wondering if the squeeze rule would be relevant here? but I do not know how to derive g(x) and h(x) from the above so that I can show g(x) ≤ f(x) ≤ h(x)

Once I know that then I think I can continue. I can already see that if f(0) then both would = 3
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 3, 2012

### Simon Bridge

Here, let me:$$f(x)=\left \{ \begin{array}{rl} 2x^2+x+3, & x<0\\ \\ \frac{3}{x+1}, & x \geq 0 \end{array}\right .$$

Why not just check that the left and right hand limits are the same?
http://www.sparknotes.com/math/calcab/functionslimitscontinuity/section3.rhtml

Last edited: Jul 3, 2012
3. Jul 3, 2012

### SammyS

Staff Emeritus
No need for the squeeze theorem.

Find the limit as x approaches zero from the right, then from the left. Are those limits equal? Do they equal f(0) ?

One of many ways to display the function in LaTeX:

$\displaystyle f(x)=\left\{\matrix{2x^2+x+3,\ \ x<0\\ \\ \frac{3}{x+1},\ \ x\ge0 }\right.$

The LaTeX code:
Code (Text):
$\displaystyle f(x)=\left\{\matrix{2x^2+x+3,\ \ x<0 \\ \\ \frac{3}{x+1},\ \ x\ge0 }\right.$

4. Jul 3, 2012

### fireychariot

All I have written then is this

xn → 0 $\Rightarrow$ 2xn+xn+3 → 3
where xn < 0

and

xn→ 0 $\Rightarrow$ $\frac{3}{x+1}$ → 3
where xn≥0

This holds by the combination rules for sequences therefore f is continuous at 0.

Is this all I have to write as it doesn't seem to be enough??

5. Jul 3, 2012

### Pranav-Arora

I don't see any xn in the question.

Well that seems correct if you mean this:
$$\lim_{x→0^-} 2x^2+x+3=3$$
$$\lim_{x→0^+} \frac{3}{x+1}=3$$

Yes, its enough to show that if the left and right hand limits are equal, the function is continuous.

6. Jul 3, 2012

### LCKurtz

Don't forget you also have to observe that f(0)=3.

7. Jul 3, 2012