Question

I need help learning limits....

For the given function, ƒ, determine if the function is discontinuous at a given point. State which condition of continuity fails or state the function is continuous at x = a.

f(x)= {
((x^2-9)/(x+3)), x<-3
x-9, x=-3
-6, x>-3

a=-3
(hopefully this makes sense)

Answer 1

We all need to learn limits!!

The basic rule for denominators is three-fold:
1) If it can be zero, it's NOT in the Domain, and
2) The function likely shows asymptotic behavior, there UNLESS
3) The numerator is also zero, there, but it's still not in the Domain.

You have \(\dfrac{x^2 - 9}{x+3}\). This expression is EXACTLY the same as x-3 EXCEPT at x = -3. Since x = -3 is not in the Domain of this expression, we can just simplify to x-3.

This leaves:

\[ f(x) = \left\{
\begin{array}{l l}
x-3 & \quad \mbox{if x < -3}\\
x-9 & \quad \mbox{if x = -3}\\
-6 & \quad \mbox{if x > -3}\\
\end{array} \right. \]

It looks like right around x = -3 is your ONLY concern. Ask yourself three questions...

1) What is the limit of this function as we approach x = -3 from the left?
2) What is the value of the function defined at x = -3?
3) What is the limit of this function as we approach x = -3 from the right?

If it's continuous, they had better be the same answer.

Answer 2

a function f(x) is continuous at x=a if
\[\lim_{x \rightarrow a}f \left( x \right)=f \left( a \right)\]

Answer 3

so from here can I just graph it to see where the functions lay on x=-3?

Answer 4

No, you cannot do that. AT x = -3, it is defined for only one segment. The other two segments require LIMIT considerations. Remember, we're learning limits, here.

Answer 5

\[\lim_{x \rightarrow -3-}f(x)=\lim_{x \rightarrow -3-}\frac{ x^2-9 }{ x+3 }=\lim_{x \rightarrow -3-}\left( x-3 \right)=-3-3=-6\]

Answer 6

similarly you \[for \lim_{x \rightarrow -3+}f(x)=\lim_{x \rightarrow -3+}(-6)=?\]
then find f(-3) and see if all are equal

Answer 7

oh okay, so because its approaching -6 from both the left and the right, the limit exists at -6? but now we need to find out if it is a real number. Because the middle function is equal to -12 that means they are not all equal.

am i on the right track?

Answer 8

so that means its not continuous correct? because \[f(a)\neq(-6) \]

Answer 9

(-3)**

Answer 10

\[f(-3)=-3-9=-12\]
\[\lim_{x \rightarrow -3}f(x) \neq f(-3)\]
so f(x) is not continuous at x=-3

Answer 11

awesome thank you so very much for the help!!

Answer 12

yw