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Question

Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin to the power of negative 1 end exponent space x over denominator x end fraction

  1. 1
  2. 0
  3. 2
  4. does not exist

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 0 end subscript fraction numerator sin to the power of negative 1 end exponent space x over denominator x end fraction.

The correct answer is: 1


    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin to the power of negative 1 end exponent space x over denominator x end fraction
    We first try substitution:
    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin to the power of negative 1 end exponent space x over denominator x end fraction = Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin to the power of negative 1 end exponent space 0 over denominator 0 end fraction 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 0 end subscript fraction numerator sin to the power of negative 1 end exponent space x over denominator x end fraction  ( L'Hopital's Rule for zero over zero ;Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator f open parentheses x close parentheses over denominator g left parenthesis x right parenthesis end fraction space equals space fraction numerator f apostrophe open parentheses 0 close parentheses over denominator g apostrophe left parenthesis 0 right parenthesis end fraction )
    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator square root of open parentheses 1 minus x squared close parentheses end root end fraction    (The derivative of sin to the power of negative 1 end exponent space x is fraction numerator 1 over denominator √ left parenthesis 1 minus x squared right parenthesis end fraction, where -1 < x < 1)
    On substituting, We get
    Lt subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator square root of open parentheses 1 minus x squared close parentheses end root end fraction = 1

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

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