lim as x approaches 0 of the [(cube root of x+1) -1] / [x]?
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Sorry, I read that as (x+1)^3. Anyway, for the lim [(x+1)^(1/3)-1]/x as x-->0, use L'Hospital's rule and take the derivative of the numerator and the derivative of the denominator:
f (x) = [(x+1)^(1/3)-1] --> f ' (x) = (1/3)(x+1)^(-2/3)
g (x) = x --> g' (x) = 1
so that
=lim [(1/3)(x+1)^(-2/3)/1] as x-->0,
=1/3
f (x) = [(x+1)^(1/3)-1] --> f ' (x) = (1/3)(x+1)^(-2/3)
g (x) = x --> g' (x) = 1
so that
=lim [(1/3)(x+1)^(-2/3)/1] as x-->0,
=1/3
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lim (³√x+1 - 1) / x ==H==> (1/3³√(x+1)²) / 1 = (1/3³√1²) / 1 = 1/3
x ->1
* d/dx(³√x+1 - 1) = 1 / 3³√(x+1)²
* x' = 1
x ->1
* d/dx(³√x+1 - 1) = 1 / 3³√(x+1)²
* x' = 1