Maths-
General
Easy

Question

If f left parenthesis 9 right parenthesis equals 9 comma f to the power of 1 left parenthesis 9 right parenthesis minus 4 comma then L t divided by left parenthesis x rightwards arrow t right parenthesis space space left parenthesis √ left parenthesis f left parenthesis x right parenthesis right parenthesis minus 3 right parenthesis divided by left parenthesis √ x minus 3 right parenthesis

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

hintHint:

In this question we are getting 0 over 0 form. So, we will use L'Hôpital's rule to find the limit.

The correct answer is: 4


    In this question we are given f left parenthesis 9 right parenthesis equals 9,f apostrophe left parenthesis 9 right parenthesis equals 4 and we have to find the limit of limit as x minus greater than 9 of fraction numerator square root of f left parenthesis x right parenthesis end root minus 3 over denominator square root of x minus 3 end fraction
    Step1: Putting the value of limit
    By putting the value of limits in the expression, we are getting both numerator and denominator as zero. That is 0 over 0 form.
    Step2: Using L'Hôpital's rule
    In this rule we differentiate both numerator and denominator.
    => limit as x minus greater than 9 of fraction numerator begin display style fraction numerator f apostrophe left parenthesis x right parenthesis over denominator 2 square root of f left parenthesis x right parenthesis end root end fraction end style over denominator begin display style fraction numerator 1 over denominator 2 square root of x end fraction end style end fraction
    By putting the value of limit we get,
    f apostrophe left parenthesis 9 right parenthesis equals 4
    Hence, the value of limit is 4.

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