CBSE Class 11-science Answered

$Find space the space value space of space limit as straight x rightwards arrow 1 of open parentheses 2 minus straight x close parentheses to the power of tan πx over 2 end exponent$

Asked by Topperlearning User | 25 Jul, 2014, 15:51: PM

Expert Answer

$straight e to the power of limit as straight x rightwards arrow 1 of tan πx over 2 log open parentheses 2 minus straight x close parentheses end exponent equals straight e to the power of limit as straight x rightwards arrow 1 of tan πx over 2 log open parentheses 1 plus stack 1 minus straight x with bar on top close parentheses end exponent Let space 1 minus straight x equals straight z comma space if space straight x rightwards arrow 1 comma space straight z rightwards arrow 0 equals straight e to the power of blank to the power of limit as straight z rightwards arrow 0 of tan fraction numerator straight pi left parenthesis 1 minus straight z right parenthesis over denominator 2 end fraction log open parentheses 1 plus straight z close parentheses end exponent end exponent equals straight e to the power of limit as straight z rightwards arrow 0 of tan fraction numerator straight pi left parenthesis 1 minus straight z right parenthesis over denominator 2 end fraction limit as straight z rightwards arrow 0 of fraction numerator log open parentheses 1 plus straight z close parentheses over denominator straight z end fraction. straight z end exponent equals straight e to the power of limit as straight z rightwards arrow 0 of tan fraction numerator straight pi left parenthesis 1 minus straight z right parenthesis over denominator 2 end fraction.1. straight z space space space space space space space space space space space space space left square bracket limit as straight z rightwards arrow 0 of fraction numerator log open parentheses 1 plus straight z close parentheses over denominator straight z end fraction equals 1 right square bracket end exponent equals e to the power of limit as z rightwards arrow 0 of tan open parentheses pi over 2 minus fraction numerator pi z over denominator 2 end fraction close parentheses.1. z end exponent equals e to the power of limit as z rightwards arrow 0 of c o t open parentheses fraction numerator pi z over denominator 2 end fraction close parentheses. z end exponent equals e to the power of limit as z rightwards arrow 0 of fraction numerator z over denominator tan begin display style fraction numerator pi z over denominator 2 end fraction end style end fraction end exponent equals e to the power of limit as z rightwards arrow 0 of fraction numerator begin display style fraction numerator pi z over denominator 2 end fraction end style over denominator tan begin display style fraction numerator pi z over denominator 2 end fraction end style end fraction.2 over pi end exponent equals e to the power of 1.2 over pi space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end exponent open square brackets because limit as z rightwards arrow 0 of fraction numerator begin display style fraction numerator pi z over denominator 2 end fraction end style over denominator tan begin display style fraction numerator pi z over denominator 2 end fraction end style end fraction equals 1 close square brackets equals e to the power of 2 over pi end exponent$

Answered by | 25 Jul, 2014, 17:51: PM