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Asked by sarveshvibrantacademy | 17 May, 2019, 09:18: AM

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begin mathsize 16px style According space to space the space question space straight f left parenthesis straight x right parenthesis space is space continuous space at space straight x equals 0. limit as straight x rightwards arrow 0 to the power of plus of space straight f left parenthesis straight x right parenthesis equals straight b limit as straight x rightwards arrow 0 of space open parentheses 1 plus ax close parentheses to the power of 1 over straight x end exponent equals straight b limit as straight x rightwards arrow 0 of space open parentheses 1 plus ax close parentheses to the power of 1 over ax cross times straight a end exponent equals straight b straight e to the power of straight a equals straight b straight a equals log space straight b..... left parenthesis straight i right parenthesis limit as straight x rightwards arrow 0 to the power of minus of space straight f left parenthesis straight x right parenthesis equals straight b limit as straight x rightwards arrow 0 of space fraction numerator open parentheses straight x plus straight c close parentheses to the power of begin display style 1 third end style end exponent minus 1 over denominator open parentheses straight x plus straight c close parentheses to the power of 1 half end exponent minus 1 end fraction equals straight b Put space straight x plus straight c equals straight h straight x rightwards arrow 0 space then space straight h rightwards arrow straight c limit as straight h rightwards arrow straight c of space fraction numerator straight h to the power of begin display style 1 third end style end exponent minus 1 over denominator straight h to the power of 1 half end exponent minus 1 end fraction equals straight b limit as straight h rightwards arrow straight c of space fraction numerator begin display style fraction numerator straight h to the power of begin display style 1 third end style end exponent minus 1 over denominator straight h minus 1 end fraction end style over denominator begin display style fraction numerator straight h to the power of 1 half end exponent minus 1 over denominator straight h minus 1 end fraction end style end fraction equals straight b fraction numerator begin display style 1 third end style over denominator begin display style 1 half end style end fraction equals straight b 2 over 3 equals straight b..... left parenthesis ii right parenthesis Find space straight a space and space straight b space from space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis. space end style

Answered by Sneha shidid | 17 May, 2019, 10:20: AM

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