Maths-
General
Easy

Question

Lt subscript x not stretchy rightwards arrow straight infinity end subscript space open parentheses square root of x squared plus x end root minus x close parentheses

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

hintHint:

In this question, we have to find value of Lt subscript x not stretchy rightwards arrow straight infinity end subscript space open parentheses square root of x squared plus x end root minus x close parentheses.

The correct answer is: 1 half


    Lt subscript x not stretchy rightwards arrow straight infinity end subscript space open parentheses square root of x squared plus x end root minus x close parentheses
    The initial form for the limit is indeterminate ∞ - ∞ So, use the conjugate
    space open parentheses square root of x squared plus x end root minus x close parentheses fraction numerator square root of x squared plus x end root plus x over denominator square root of x squared plus x end root plus x end fraction space equals fraction numerator x squared plus x minus x squared over denominator square root of x squared plus x end root plus x end fraction space equals space fraction numerator x over denominator square root of x squared plus x end root plus x end fraction
    L t subscript x rightwards arrow infinity end subscript fraction numerator x over denominator square root of x squared plus x end root plus x end fraction space h a s space i n d e t e r m i n a t e space f o r m space infinity over infinity comma space b u t space w e space c a n space f a c t o r space a n d space r e d u c e.
    L t subscript x rightwards arrow infinity end subscript fraction numerator x over denominator square root of x squared plus x end root plus x end fraction space equals L t subscript x rightwards arrow infinity end subscript fraction numerator x over denominator x square root of 1 plus begin display style 1 over x end style end root plus x end fraction
    Also we can write ,
    L t subscript x rightwards arrow infinity end subscript fraction numerator 1 over denominator square root of 1 plus begin display style 1 over x end style end root plus 1 end fraction
    On substituting, We get
    fraction numerator 1 over denominator square root of 1 plus begin display style 1 over infinity end style end root plus 1 end fraction space equals space 1 half

    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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