1. ## continuous function

k(x) is a piecewise function

if k(x)= (x^2-2b+b^2)/(x-b) if x does not equal a
= 5 if x=a

then which of the following is true about k(x):
1) limx->a k(x) exists
2) k(x) is continuous at x=a
3) k(a) exists

so far i think that k(a) will exist because it will equal 5 right? realli not sure about the other two.

2. Hi chukie! First, let me make your notation easier to read:

Originally Posted by chukie
$\displaystyle k(x)$ is a piecewise function

$\displaystyle k(x) = \left\{\begin{array}{rl} \frac{x^2-2b+b^2}{x-b}, & \text{if }x\neq a\\ 5, & \text{if }x=a \end{array}\right.$

then which of the following is true about $\displaystyle k(x)$:

1) $\displaystyle \lim_{x\to a}k(x)\text{ exists}$

2) $\displaystyle k(x) \text{ is continuous at }x=a$

3) $\displaystyle k(a) \text{ exists}$

so far i think that $\displaystyle k(a)$ will exist because it will equal 5 right? realli not sure about the other two.
You are correct that $\displaystyle k$ is defined at $\displaystyle a$.

For the others:

1. When taking a limit, we only care about what happens to a function as it approaches a certain value, not what happens at that value. So, the limit of $\displaystyle k$ as $\displaystyle x\to a$ will be the same as the limit of the first "piece."

2. For a function $\displaystyle f(x)$ to be continuous at a point $\displaystyle x = c$, three things must be true:

* $\displaystyle f(c)\text{ is {d}efined}$

* $\displaystyle \lim_{x\to c}f(x) \text{ exists}$

* $\displaystyle \lim_{x\to c}f(x) = f(c)$

You should know the first 2 by answering the other two parts of this problem. So, take the limit of $\displaystyle k(x)$ as $\displaystyle x\to a$. Then, if the value of this limit is $\displaystyle k(a) = 5$, then $\displaystyle k$ is continuous at $\displaystyle a$, and it will be discontinuous otherwise.

You should be able to find which value for $\displaystyle b$ makes the function continuous.

3. thanks so much! ur awesome