Maths-
General
Easy

Question

For x element of R L t left parenthesis x rightwards arrow infinity right parenthesis left parenthesis left parenthesis x minus 3 right parenthesis divided by left parenthesis x plus 2 right parenthesis right parenthesis to the power of x equals

  1. e
  2. e minus 1
  3. e-5
  4. e5

hintHint:

In this question we are getting 1 to the power of infinity form. We will use Standard limits formula limit as x rightwards arrow infinity of f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent equals space e to the power of limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis minus 1 right parenthesis g left parenthesis x right parenthesis end exponent to find the limit

The correct answer is: e-5


    In this question we have to find the limit of limit as x rightwards arrow infinity of open parentheses fraction numerator x minus 3 over denominator x plus 2 end fraction close parentheses to the power of x
    Step1: Rearranging the equation
    By dividing by x in both numerator and denominator we get,
    limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x end style over denominator 1 plus begin display style 2 over x end style end fraction close parentheses to the power of x
    By putting the value of limit we are getting 1 to the power of infinity form.
    Step2: Using Standard limits
    We know that limit as x rightwards arrow infinity of f left parenthesis x right parenthesis to the power of g left parenthesis x right parenthesis end exponent equals space e to the power of limit as x rightwards arrow infinity of left parenthesis f left parenthesis x right parenthesis minus 1 right parenthesis g left parenthesis x right parenthesis end exponent
    e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x end style over denominator 1 plus begin display style 2 over x end style end fraction minus 1 close parentheses x end exponent
    => e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator 1 minus begin display style 3 over x minus 1 minus 2 over x end style over denominator 1 plus begin display style 2 over x end style end fraction close parentheses x end exponent
    => e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 over x end style over denominator begin display style fraction numerator 2 plus x over denominator x end fraction end style end fraction close parentheses x end exponent
    => e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 x end style over denominator begin display style x plus 2 end style end fraction close parentheses end exponent
    =>e to the power of limit as x rightwards arrow infinity of open parentheses fraction numerator negative begin display style 5 end style over denominator begin display style 1 plus fraction numerator begin display style 2 end style over denominator x end fraction end style end fraction close parentheses end exponent
    By putting the value of limits we get
    e to the power of negative 5 end exponent
    Hence, the value of limit us e to the power of negative 5 end exponent.

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