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Question

Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator cube root of x minus 1 over denominator x minus 1 end fraction

  1. 2 over 3
  2. 3 over 2
  3. 1 third
  4. 1 half

hintHint:

We can apply L'Hopital's rule, also commonly spelled L'Hospital's rule, whenever direct substitution of a limit yields an indeterminate form. This means that the limit of a quotient of functions (i.e., an algebraic fraction) is equal to the limit of their derivatives.
In this question, we have to find value of Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator cube root of x minus 1 over denominator x minus 1 end fraction.

The correct answer is: 2 over 3


    Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator cube root of x minus 1 over denominator x minus 1 end fraction
    We first try substitution:
    Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator cube root of x minus 1 over denominator x minus 1 end fraction = Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator cube root of 1 minus 1 over denominator 1 minus 1 end fraction space equals space 0 over 0
    Since the limit is in the form 0 over 0, it is indeterminate we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit.

    Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator left parenthesis x right parenthesis to the power of begin display style bevelled 1 third end style end exponent minus 1 to the power of begin display style bevelled 1 third end style end exponent over denominator x minus 1 end fraction    (We know that ,Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator x to the power of n minus a to the power of n over denominator x minus a end fraction space equals space n a to the power of n minus 1 end exponent)
    Therefore we can simply write,
    Lt subscript x not stretchy rightwards arrow 1 end subscript space fraction numerator left parenthesis x right parenthesis to the power of begin display style bevelled 1 third end style end exponent minus 1 to the power of begin display style bevelled 1 third end style end exponent over denominator x minus 1 end fraction = 1 third space cross times space 1 to the power of 1 third minus 1 end exponent equals space 1 third

    We can only apply the L’Hospital’s rule if the direct substitution returns an indeterminate form, that means 0 over 0 space o r space fraction numerator plus-or-minus infinity over denominator plus-or-minus infinity end fraction.

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