Answer:
use l'Hopital's rule
Step-by-step explanation:
The expression given evaluates to 0/0 at x=1, so l'Hopital's rule can be used to find the limit. This requires you to differentiate the numerator and denominator separately and find the ratio of the derivatives at x=1.
n = √(2x -1) +x^2 -3x +1
n' = 1/√(2x -1) +2x -3 ⇒ 1/√(2-1) +2 -3 = 0 . . . at x=1
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d = ∛(x -2) +x^2 -x +1
d' = 1/(3∛((x-2)^2)) +2x -1 ⇒ 1/(3∛1) +2 -1 = 4/3 . . . at x=1
Then the limit is ...
n'/d' = 0/(4/3) = 0 . . . . limit as x approaches 1
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A graphing or spreadsheet calculator can be helpful. The attached graph is expected to have a hole at x=1, but does not. Whereas numerator and denominator are both zero at x=1, the calculators seems perfectly happy to evaluate their ratio there.
