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General
Easy

Question

L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus sin space x minus 1 over denominator x end fraction

  1. 1
  2. e to the power of x
  3. x
  4. 0

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 L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus sin space x minus 1 over denominator x end fraction.

The correct answer is: 0


    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus sin space x minus 1 over denominator x end fraction
    We first try substitution:
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus sin space x minus 1 over denominator x end fraction = L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of 0 minus sin space 0 minus 1 over denominator 0 end fraction = 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.
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus sin space x minus 1 over denominator x end fraction =  L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus 1 over denominator x end fraction-L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin space space x over denominator x end fraction   (We know that L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus 1 over denominator x end fraction = 1 &  L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin space space x over denominator x end fraction=1.)
    So, L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator e to the power of x minus 1 over denominator x end fractionL t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator sin space space x over denominator x end fraction = 1- 1 = 0

    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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