Let h(x) =
(
x2 + 2 if x > 2
��mx if x � 2
(a) Find the value of m that will make h continuous everywhere.
(b) Using the value of m you have found to determine limx!2 h(x)
(a) To make h continuous everywhere, we need the left-hand limit and the right-hand limit of h(x) to be equal at x = 2.
First, let's find the left-hand limit of h(x) as x approaches 2:
lim(x->2-) h(x) = lim(x->2-) mx = 2m
Next, let's find the right-hand limit of h(x) as x approaches 2:
lim(x->2+) h(x) = lim(x->2+) (x^2 + 2) = 2^2 + 2 = 6
For the function to be continuous at x = 2, the left-hand limit and the right-hand limit should be equal:
2m = 6
Solving this equation for m:
2m = 6
m = 6/2
m = 3
So, the value of m that will make h continuous everywhere is m = 3.
(b) Now that we have the value of m, let's determine the limit of h(x) as x approaches 2.
lim(x->2) h(x) can be found by calculating the value of h(x) for x values approaching 2 from both sides.
For x values less than 2 (x -> 2-), h(x) = mx = 3x.
lim(x->2-) h(x) = lim(x->2-) 3x = 3(2) = 6
For x values greater than 2 (x -> 2+), h(x) = x^2 + 2.
lim(x->2+) h(x) = lim(x->2+) (x^2 + 2) = 2^2 + 2 = 6
Since the left-hand limit and the right-hand limit are equal, we can conclude that:
lim(x->2) h(x) = 6.