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Question

text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript left parenthesis square root of x plus 1 end root minus square root of x right parenthesis

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

hintHint:

In this question, we have to find value of text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript left parenthesis square root of x plus 1 end root minus square root of x right parenthesis.

The correct answer is: 0


    text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript left parenthesis square root of x plus 1 end root minus square root of x right parenthesis
    The initial form for the indeterminate infinity minus infinity. So use conjugate,
    left parenthesis square root of x plus 1 end root minus square root of x right parenthesis fraction numerator left parenthesis square root of x plus 1 end root plus square root of x right parenthesis over denominator left parenthesis square root of x plus 1 end root plus square root of x right parenthesis end fraction space equals fraction numerator left parenthesis x plus 1 minus x right parenthesis over denominator left parenthesis square root of x plus 1 end root plus square root of x right parenthesis end fraction space equals fraction numerator 1 over denominator left parenthesis square root of x plus 1 end root plus square root of x right parenthesis end fraction
    text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript fraction numerator 1 over denominator left parenthesis square root of x plus 1 end root plus square root of x right parenthesis end fraction
    Also we can write ,
    text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript fraction numerator 1 over denominator square root of x left parenthesis square root of 1 plus begin display style 1 over x end style end root plus 1 right parenthesis end fraction text  =  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript fraction numerator 1 over denominator square root of x end fraction text × Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript fraction numerator 1 over denominator left parenthesis square root of 1 plus begin display style 1 over x end style end root plus 1 right parenthesis end fraction
    On substituting, We get
    text  Lt  end text subscript x not stretchy rightwards arrow straight infinity end subscript fraction numerator 1 over denominator square root of x left parenthesis square root of 1 plus begin display style 1 over x end style end root plus 1 right parenthesis end fraction text = 0 end text

    We can only apply the L’Hospital’s rule if the direct substitution returns an indeterminate form, that means 0 over 0 or infinity over infinity.

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