Maths-
General
Easy

Question

L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator square root of x plus 1 end root minus 1 over denominator x end fraction

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

hintHint:

If the function has a square root in it and Substitution yields 0 over 0Error converting from MathML to accessible text., 0 divided by 0, then multiply the numerator and the denominator by  
1=Error converting from MathML to accessible text.
As with Factoring, this approach will probably lead to being able to cancel a term.

The correct answer is: 1 half


    We first try substitution:
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator square root of x plus 1 end root minus 1 over denominator x end fraction = fraction numerator square root of 0 plus 1 end root minus 1 over denominator 0 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.

    So let’s get rid of the square roots, using the conjugate just like you practiced in algebra: multiply both the numerator and denominator by the conjugate of the numerator square root of x plus 1 space end root plus 1square root of blank.
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator square root of x plus 1 end root minus 1 over denominator x end fraction
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator square root of x plus 1 end root minus 1 over denominator x end fractionfraction numerator square root of x plus 1 space end root space plus 1 over denominator square root of x plus 1 space end root space plus 1 end fraction
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator left parenthesis square root of x plus 1 end root space right parenthesis squared minus 1 over denominator x left parenthesis square root of x plus 1 end root space plus 1 right parenthesis end fraction
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator x over denominator x left parenthesis square root of x plus 1 end root space plus 1 right parenthesis end fraction
    L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator 1 over denominator left parenthesis square root of x plus 1 end root space plus 1 right parenthesis end fraction
    Put the value x = 0 so, fraction numerator 1 over denominator 1 plus 1 end fraction = 1 half
    Therefore limit value of  L t subscript x not stretchy rightwards arrow 0 end subscript fraction numerator square root of x plus 1 end root minus 1 over denominator x end fraction = 1 half
     

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