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Question

Lt subscript x not stretchy rightwards arrow straight infinity end subscript open square brackets square root of x squared plus a x plus b end root minus x close square brackets

  1. a
  2. 2a
  3. 1
  4. a over 2

hintHint:

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

The correct answer is: a over 2


    Lt subscript x not stretchy rightwards arrow straight infinity end subscript open square brackets square root of x squared plus a x plus b end root minus x close square brackets
    The initial form the limit is indeterminate infinity minus infinity .So, use the conjugate
    square root of x squared plus a x plus b end root minus x space cross times fraction numerator square root of x squared plus a x plus b end root plus x over denominator square root of x squared plus a x plus b end root plus x end fraction space equals space fraction numerator x squared plus a x plus b minus x squared over denominator square root of x squared plus a x plus b end root plus x end fraction
    L t subscript x rightwards arrow infinity end subscript fraction numerator a x plus b over denominator square root of x squared plus a x plus b 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.
    A l s o space W e space c a n space w r i t e space t h i s space w a y space L t subscript x rightwards arrow infinity end subscript fraction numerator a plus begin display style b over x end style over denominator square root of 1 plus begin display style a over x end style plus begin display style b over x squared end style end root plus 1 end fraction    (We know that square root of x squared end root equals open vertical bar x close vertical bar comma comma space s o space f o r space p o s i t i v e space x)
    On substituting, We get
    L t subscript x rightwards arrow infinity end subscript fraction numerator a plus b over x over denominator square root of 1 plus begin display style a over x end style plus begin display style b over x squared end style end root plus 1 end fraction space equals space a over 2

    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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