# Thread: Limit help ? Limit Continuous ? part 1

1. ## Limit help ? Limit Continuous ? part 1

Can someone solve for this problem ? Thank you so much !!!

1. Define h(2) so that $h(x)=( X^2 + 3X - 10 )/( X-2 )$ IS CONTINUOUS AT X =1

2. wHAT VALUE SHOULD BE ASSIGNED to b to make the function

g (x) = { x^3 , x < 1/2
{ bx^2 ,x > and equal 1/2

3. Explain why the limits exist and give their values . Support with graphing utility !!

$lim [ ( 1 + cos X ) / 2 ]$
X--> 0

Thank you !!!

2. #2:

$g(1/2)=\frac{1}{8}$

Find b so that $\lim_{x\to \frac{1}{2}^{+}}bx^{2}=\frac{1}{8}$

3. Can you factor $x^2 + 3x – 10$?
If so, then if you do not understand at that point, you are really over you head!
$\frac{{x^2 + 3x - 10}}{{x - 2}} = \frac{{\left( {x - 2} \right)\left( {x + 5} \right)}}{{x - 2}} = x + 5\,,\,x \ne 2$.

4. ## yes ...!!

yes I understand ..that part ..I just need to know how to we define ..the continuous at x=2 !!!

If I 've said anything wrong ..I'm sorry about that !!!

5. Originally Posted by mathfun9
yes I understand ..that part ..I just need to know how to we define ..the continuous at x=2 !!!

If I 've said anything wrong ..I'm sorry about that !!!
Find the limit as x approaches 2. Let's name it L. Then define a piece-wise function, such that:

$\left\{ \begin{array}{rcl} \frac{x^2+3x-10}{x-2} & x\neq2 \\
L & x = 2 \end{array} \right.$