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CBSE Class 11-science Answered

Without use of L.hospital rule
question image
Asked by rahoriya72 | 18 Mar, 2020, 10:53: AM
answered-by-expert Expert Answer
limit as straight x rightwards arrow 0 of fraction numerator straight x space log space cosx over denominator log open parentheses 1 plus straight x squared close parentheses end fraction
equals limit as straight x rightwards arrow 0 of fraction numerator straight x space log space cosx over denominator straight x squared cross times begin display style fraction numerator log open parentheses 1 plus straight x squared close parentheses over denominator straight x squared end fraction end style end fraction
equals fraction numerator limit as straight x rightwards arrow 0 of 1 over straight x cross times log space cosx over denominator begin display style 1 end style end fraction..... open parentheses limit as straight x rightwards arrow 0 of fraction numerator log open parentheses 1 plus straight x squared close parentheses over denominator straight x squared end fraction equals 1 close parentheses
numerator
limit as straight x rightwards arrow 0 of 1 over straight x cross times log space cosx
equals limit as straight x rightwards arrow 0 of 1 over straight x cross times log space open parentheses 1 minus fraction numerator straight x squared over denominator 2 factorial end fraction plus fraction numerator straight x to the power of 4 over denominator 4 factorial end fraction minus..... close parentheses 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 space... expansion space of space cos space straight x
equals limit as straight x rightwards arrow 0 of 1 over straight x cross times log space open parentheses 1 plus open parentheses negative fraction numerator straight x squared over denominator 2 factorial end fraction plus fraction numerator straight x to the power of 4 over denominator 4 factorial end fraction minus..... close parentheses close parentheses space
equals limit as straight x rightwards arrow 0 of 1 over straight x cross times open parentheses open parentheses negative fraction numerator straight x squared over denominator 2 factorial end fraction plus fraction numerator straight x to the power of 4 over denominator 4 factorial end fraction minus..... close parentheses minus open parentheses negative fraction numerator straight x squared over denominator 2 factorial end fraction plus fraction numerator straight x to the power of 4 over denominator 4 factorial end fraction minus..... close parentheses squared over 2 plus.... close parentheses 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... expansion space of space log space left parenthesis 1 plus straight x right parenthesis
equals limit as straight x rightwards arrow 0 of open parentheses negative fraction numerator straight x over denominator 2 factorial end fraction plus fraction numerator straight x cubed over denominator 4 factorial end fraction minus..... close parentheses minus 1 over straight x cross times higher space powers space of space straight x to the power of straight n
applying space limits
equals 0
so space limit space is space 0 over 1 equals 0
straight f left parenthesis 0 right parenthesis equals 0... given
hence space continuous
Answered by Arun | 19 Mar, 2020, 05:22: PM

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