Answer:
1. lim x→0 7e^x =7
2. lim f(x) does not exist
f(3) = 5
3. 1/12
Explanation:
f(x) = 7e^x
lim x→0 7e^x =7
AS we can see from the graph it is around 7
We know that e^0 =1 so 7*1 =7
2. The limit does not exists, because approaching the limit from the left f(3-) is 3. Approaching the limit from the right f(3+) is 5. f(3) = 5 since there is a closed circle at 3 with a value of 5.
3. f(x) = x^ 1/2
We take the derivative
f'(x) = 1/2 x^ (-1/2)
= 1/2 1/( x^.5)
To find it at x=36
= 1/2 * 1/( 36^.5)
= 1/2 * 1/6
= 1/12
Since they are not equal, the limit does not exist.