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General
Easy

Question

Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction

  1. 1
  2. 0
  3. 2
  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 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction.

The correct answer is: 2


    Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction
    We first try substitution :
    Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction = fraction numerator e to the power of 0 minus 1 over denominator square root of 1 plus 0 end root minus 1 end fraction = fraction numerator 1 minus 1 over denominator 1 minus 1 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.
    Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator square root of 1 plus x end root minus 1 end fraction    ( we knowLt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator x end fraction space equals space 1 )
    Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator e to the power of x minus 1 over denominator x end fraction cross times space Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator x over denominator square root of 1 plus x end root minus 1 end fraction
    Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator space left parenthesis 1 plus x right parenthesis space minus 1 over denominator square root of 1 plus x end root minus 1 end fraction = Lt subscript x not stretchy rightwards arrow 0 end subscript space fraction numerator space left parenthesis square root of 1 plus x end root right parenthesis squared minus 1 squared over denominator square root of 1 plus x end root minus 1 end fraction     ( Let square root of 1 plus x end root = y,  we know Lt subscript x not stretchy rightwards arrow a end subscript space fraction numerator space left parenthesis x right parenthesis squared minus a squared over denominator x space minus a end fraction space equals space n a to the power of n minus 1 end exponent )
    Now, We can write simply

    Lt subscript y not stretchy rightwards arrow 1 end subscript space fraction numerator space left parenthesis y right parenthesis squared minus 1 squared over denominator y minus 1 end fraction = 2 cross times space 1 to the power of left parenthesis 2 minus 1 right parenthesis end exponent =2

    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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