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CBSE Class 11-science Answered

Asked by Topperlearning User | 25 Jul, 2014, 13:08: PM
answered-by-expert Expert Answer

limit as x rightwards arrow pi over 2 of fraction numerator sin left parenthesis cos space x right parenthesis space cos space x over denominator sin space x minus cos e c space x end fraction
equals limit as x rightwards arrow pi over 2 of fraction numerator sin left parenthesis cos space x right parenthesis space cos squared space x over denominator cos space x open parentheses sin space x minus begin display style fraction numerator 1 over denominator sin space x end fraction end style close parentheses end fraction
equals limit as x rightwards arrow pi over 2 of fraction numerator sin left parenthesis cos space x right parenthesis space over denominator cos space x end fraction. limit as x rightwards arrow pi over 2 of fraction numerator cos squared space x. sin space x over denominator open parentheses sin squared space x minus 1 close parentheses end fraction space space space space space space space space space space space space space space space space open square brackets limit as x rightwards arrow 0 of fraction numerator sin space x over denominator x end fraction equals 1 close square brackets
equals limit as cos space x rightwards arrow 0 of fraction numerator sin left parenthesis cos space x right parenthesis space over denominator cos space x end fraction. limit as x rightwards arrow pi over 2 of fraction numerator left parenthesis 1 minus sin squared space x right parenthesis. sin space x over denominator minus open parentheses 1 minus sin squared space x close parentheses end fraction
equals 1. space limit as x rightwards arrow pi over 2 of minus sin space x
equals minus 1. space sin space pi over 2 equals minus 1

Answered by | 25 Jul, 2014, 15:08: PM
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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
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