No l'hospital rule involved please
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Use the following difference of cubes formula
(a^3 - b^3) = (a - b)(a² + ab + b²).
Using x^(1/3) = a and 2^(1/3) = b gives
(x - 2) = (x^(1/3) - 2^(1/3))[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)].
Now, divide both sides by x - 2 and by x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3), to get
(x^(1/3) - 2^(1/3))/(x - 2) = 1/[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)].
Since these two expressions are equal, we can use the one on the right to take the limit and avoid the 0/0 form.
lim (x^(1/3) - 2^(1/3))/(x - 2) = lim 1/[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)] = 1/(3•2^(1/3)).
x->2 . . . . . . . . . . . . . . . . . x->2
This method can be done by taking the original expression (x^(1/3) - 2^(1/3))/(x - 2) and multiplying and dividing by x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3). It's like multiplying and dividing by a conjugate.
(a^3 - b^3) = (a - b)(a² + ab + b²).
Using x^(1/3) = a and 2^(1/3) = b gives
(x - 2) = (x^(1/3) - 2^(1/3))[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)].
Now, divide both sides by x - 2 and by x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3), to get
(x^(1/3) - 2^(1/3))/(x - 2) = 1/[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)].
Since these two expressions are equal, we can use the one on the right to take the limit and avoid the 0/0 form.
lim (x^(1/3) - 2^(1/3))/(x - 2) = lim 1/[x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3)] = 1/(3•2^(1/3)).
x->2 . . . . . . . . . . . . . . . . . x->2
This method can be done by taking the original expression (x^(1/3) - 2^(1/3))/(x - 2) and multiplying and dividing by x^(2/3) + x^(1/3)2^(1/3) + 2^(2/3). It's like multiplying and dividing by a conjugate.