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Question

L subscript t not stretchy rightwards arrow 2 end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction

  1. -1
  2. 1
  3. 0
  4. 2

hintHint:

The absolute value function  | x-2 | can be defined as the piecewise function. In this question, we have to find value of
L subscript x not stretchy rightwards arrow 2 end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction.

The correct answer is: -1


    L subscript x not stretchy rightwards arrow 2 end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction
    The absolute value function  | x-2 | can be defined as the piecewise function.
    | x-2 |=  left curly bracket space space x minus 2 space space space space space space space space space semicolon space space x greater or equal than 2 space
left curly bracket space minus left parenthesis x minus 2 right parenthesis space space space semicolon space x less than 2
    we should determine if the limit from the left approaches the limit from the right .
    When the function is directly to the left of x = - 2 we are on the - (x + 2) portion of the piecewise function since x < - 2.
    Thus, the function when x<-2 becomes
    fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction equals fraction numerator negative left parenthesis x minus 2 right parenthesis over denominator x minus 2 end fraction
    Hence the limit from the left is,
    L subscript x not stretchy rightwards arrow 2 end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction = L subscript x not stretchy rightwards arrow 2 end subscript space minus 1 = -1
    From the right ,x greater than negative 2 so we just use x+2 in place | x+ 2 |
    L subscript x not stretchy rightwards arrow 2 end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction = L subscript x not stretchy rightwards arrow 2 end subscript space 1 = 1

    space L t subscript x not stretchy rightwards arrow 2 to the power of minus end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction space i s space d o e s space n o t space e x i s t. space s o
    L t subscript x not stretchy rightwards arrow 2 to the power of plus end subscript fraction numerator vertical line x minus 2 vertical line over denominator x minus 2 end fraction space equals space 1

    The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign.

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