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    Maths-
    General
    Easy

    Question

    L i m left parenthesis x rightwards arrow 0 right parenthesis left parenthesis x t a n 2 x minus 2 x t a n x right parenthesis divided by left parenthesis left parenthesis 1 minus c o s invisible function application 2 x right parenthesis squared right parenthesis equals

    1. 2
    2. -2
    3. 1 divided by 2
    4. negative 1 divided by 2

    hintHint:

    In this question first we have to use trigonometric formula   tan open parentheses 2 x close parentheses equals fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction and cos open parentheses 2 x close parentheses equals fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction and then we will use standard limits limit as x minus greater than 0 of fraction numerator x over denominator tan open parentheses x close parentheses end fraction equals 1

    The correct answer is: 1 divided by 2

      In this question we have to find the limit limit as x minus greater than 0 of fraction numerator x tan open parentheses 2 x close parentheses minus 2 x tan open parentheses x close parentheses over denominator left parenthesis 1 minus cos open parentheses 2 x close parentheses right parenthesis squared end fraction
      Step1: Using trigonometric formulae.
      we know that tan open parentheses 2 x close parentheses equals fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction and cos open parentheses 2 x close parentheses equals fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction
      => limit as x minus greater than 0 of fraction numerator x begin display style fraction numerator 2 tan open parentheses x close parentheses over denominator 1 minus tan squared open parentheses x close parentheses end fraction end style minus 2 x tan open parentheses x close parentheses over denominator open parentheses 1 minus begin display style fraction numerator 1 minus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction end style close parentheses squared end fraction
      => limit as x minus greater than 0 of fraction numerator 2 x tan open parentheses x close parentheses minus 2 x tan open parentheses x close parentheses left parenthesis 1 minus tan squared open parentheses x close parentheses right parenthesis over denominator open parentheses begin display style fraction numerator 1 plus tan squared open parentheses x close parentheses minus 1 plus tan squared open parentheses x close parentheses over denominator 1 plus tan squared open parentheses x close parentheses end fraction end style close parentheses squared end fraction
      => limit as x minus greater than 0 of fraction numerator 2 x tan cubed open parentheses x close parentheses over denominator 4 tan to the power of 4 open parentheses x close parentheses end fraction
      Step2: By using Standard limits
      We know that limit as x minus greater than 0 of fraction numerator x over denominator tan open parentheses x close parentheses end fraction equals 1
      => limit as x minus greater than 0 of fraction numerator x over denominator 2 tan open parentheses x close parentheses end fraction
      => 1 half.
      Hence, the value of limit is 1 half

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