Now, substitute the given limit:
g'(0) = Lim h--> [g(h) - g(0)]/h
= Lim h--> [1*h - 0]/h
= Lim h--> 1
= 1
Therefore, g'(0) = 1.
Suppose lim x->0 {g(x)-g(0)} / x = 1.
It follows necesarily that
a. g is not defined at x=0
b. the limit of g(x) as x approaches equals 1
c.g is not continuous at x=0
d.g'(0) = 1
The answer is d, can someone please explain how?
Thanks.
lim x->0 {g(x)-g(0)} / x = 1.
You can use the definition of the derivative:
g'(x) = Lim h--> [g(x+h) - g(x)]/h
Take x = 0:
g'(0) = Lim h--> [g(h) - g(0)]/h
And h is just a "dummy variable" whose name doesn't matter :)
1 answer