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calculus again

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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?


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 :)

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