Maths-
General
Easy

Question

L t subscript left parenthesis x rightwards arrow negative infinity right parenthesis left square bracket left parenthesis x to the power of 4 s i n invisible function application left parenthesis 1 divided by x right parenthesis plus x squared right parenthesis divided by left parenthesis left parenthesis 1 plus vertical line x vertical line cubed right parenthesis right parenthesis right square bracket equals

  1. 1
  2. -1
  3. 0
  4. negative infinity

hintHint:

In this question first we will rearrange the expression by changing open vertical bar x close vertical bar equals negative x as x will always be negative. Then we divide both denominator and numerator by x cubed. and then we use standard limits limit as x minus greater than 0 of fraction numerator sin open parentheses theta close parentheses over denominator theta end fraction equals 1.

The correct answer is: -1


    In this question we have to find the limit of limit as x rightwards arrow negative infinity of open parentheses fraction numerator x to the power of 4 sin open parentheses begin display style 1 over x end style close parentheses plus x squared over denominator 1 plus open vertical bar x close vertical bar cubed end fraction close parentheses
    Step1: Rearranging the expression.
    Since x minus greater than negative infinityx will always be negative, open vertical bar x close vertical bar equals negative x
    limit as x rightwards arrow negative infinity of open parentheses fraction numerator x to the power of 4 sin open parentheses begin display style 1 over x end style close parentheses plus x squared over denominator 1 minus x cubed end fraction close parentheses

    By dividing both numerator and denominator by x cubed, we get
    =>limit as x rightwards arrow negative infinity of open parentheses fraction numerator begin display style fraction numerator sin open parentheses begin display style 1 over x end style close parentheses over denominator begin display style 1 over x end style end fraction end style plus begin display style 1 over x end style over denominator begin display style 1 over x cubed end style minus 1 end fraction close parentheses
    Step2: Using standard limits
    We know that limit as x minus greater than 0 of fraction numerator sin open parentheses theta close parentheses over denominator theta end fraction equals 1
    =>limit as x rightwards arrow negative infinity of open parentheses fraction numerator begin display style 1 end style plus begin display style 1 over x end style over denominator begin display style 1 over x cubed end style minus 1 end fraction close parentheses
    By putting the value of limit we get
    => fraction numerator 1 plus 0 over denominator 0 minus 1 end fraction equals negative 1
    Hence, the value of the limit is negative 1.

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