Determining a functions this is continuous at 0

[fireychariot]

Homework Statement

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.

Homework Equations

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

 

[Simon Bridge]

Homework Statement

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.

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

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)

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

 

[SammyS]

Homework Statement

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.

Homework Equations

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

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:

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

 

[fireychariot]

All I have written then is this

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

and

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

 

[Saitama]

All I have written then is this

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

and

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

I don't see any x_n 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.

 

[LCKurtz]

I don't see any x_n 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.

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

 

[Saitama]

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

Oops, sorry, forgot about that.