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murillopdr
evaluate the limit of (x^(1/3)-1)/(sqrt(x)-1) as x tends to one. Algebraically.
you can apply L' hopital because you have the indetermination 0/0 and the answer is 2/3
The trick is to notice that, for the numerator: \[(x ^{1/3}-1) = [(x^{1/6})^2 - 1^2) = (x^{1/6} -1)(x^{1/6}+1)\] Also, notice that: \[\sqrt{x} = x^{1/2}=(x^{1/6})^3\] So, using a^3 - b^3 = (a - b)(a^2 + ab + b^2), the denominator is: \[[(x^{1/6})^3 - 1^3] = (x^{1/6} - 1^1)[(x^{1/6})^2 + x^{1/6}(1) + 1^2]\] Now the first term of the numerator and of the denominator cancel. Yay! \[\frac{(x^{1/6} -1)(x^{1/6}+1)}{(x^{1/6} - 1^1)[(x^{1/6})^2 + x^{1/6}(1) + 1^2]} = \frac{(x^{1/6}+1)}{[(x^{1/6})^2 + x^{1/6}(1) + 1^2]} \] Finally, substituting x= 1, will get you the result of 2/3.