By the confront theorem we know that the limit only exists if both lateral limits are equal
In this case they aren't so we don't have limit for x approaching 2, but we can find their laterals.
Approaching 2 by the left we have it on the 5 line so this limit is 5
Approaching 2 by the right we have it on the -3 line so this limit is -3
Think: it's approaching x = 2 BUT IT'S NOT 2, and we only have a different value for x = 2 which is 1, but when it's approach by the left we have the values in the 5 line and by the right in the -3 line.
Answer: $4.95
20 • .30 = 6
20 - 6 = 14
14 • .075 = 1.05
14 + 1.05 = 15.05
20 - 15.05 = 4.95
I hope this helps you
s' (t)=0-4
s' (t)=-4
t=8
s' (8)=-4
<span>Perpendicular lines' slopes are reciprocals of one another. So we know the perpendicular line will have a slope of 1/2. Then we can use our points to find the rest.
8=1/2(1)+b
8=0.5+b
7.5=b
y=2x+b</span>
