lim_{x\to 0} (27+x)^{1/3} - (27+x)^{2/3}

The correct option is C (1/6)
The given limit is in the indeterminate form 0/0
Using the L'Hospital Rule we get:
lim_{x\to 0} \frac{1}{3}(27+x)^{-2/3} - \frac{2}{3}(27+x)^{-1/3}
Now, substituting the value of x we get:
= \frac{1}{3} \times (27)^{-2/3} - \frac{2}{3} \times (27)^{-1/3}
= \frac{1}{3} \times \frac{1}{9} - \frac{2}{3} \times \frac{1}{3}
= \frac{1}{27} - \frac{2}{9} = \frac{1 - 6}{27} = -\frac{5}{27}
Let's use binomial expansion:
(27+x)^{1/3} = 27^{1/3} (1 + x/27)^{1/3} = 3(1 + x/81 + O(x^2))
(27+x)^{2/3} = 27^{2/3} (1 + x/27)^{2/3} = 9(1 + 2x/81 + O(x^2))
(27+x)^{1/3} - (27+x)^{2/3} = 3 + x/27 - 9 - 2x/27 + O(x^2) = -6 - x/27 + O(x^2)
lim_{x\to 0} ((27+x)^{1/3} - (27+x)^{2/3}) = -6
Let's try another approach:
Let f(x) = (27+x)^{1/3}
f'(x) = (1/3)(27+x)^{-2/3}
f(0) = 3
f'(0) = 1/27
Let g(x) = (27+x)^{2/3}
g'(x) = (2/3)(27+x)^{-1/3}
g(0) = 9
g'(0) = 2/9
lim_{x\to 0} \frac{f(x) - g(x)}{x} = \frac{f'(0) - g'(0)}{1} = \frac{1}{27} - \frac{2}{9} = \frac{1 - 6}{27} = -\frac{5}{27}
The question is likely incorrectly stated or has a typo.