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Compute lim( f(2+h)-f(2) )/h as h ---> 0 ?
If so, then this has the appearance of attempting to find the derivative of f(x) = x by First Principles when x = 2
Short way is to know that
f ' (x) = 1 , and so the result would be 1
long way:
lim( f(2+h)-f(2) )/h as h ---> 0
= lim ( x + h - x)/h as h --> 0
= lim h/h
= 1
Compute lim f(2+h)-f(2)
h->0 ---------
h
1 answer