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Sir pl tell the above question 

Asked by neeraj.unnao 14th March 2017, 8:49 PM
Answered by Expert
Answer:
begin mathsize 20px style limit as straight x rightwards arrow 0 of fraction numerator square root of 1 plus straight x end root minus 1 over denominator cube root of 1 plus straight x end root minus 1 end fraction
If space we space put space the space limit comma space we space will space get space 0 over 0 space form comma space so space we space use space straight L apostrophe space Hospital space rule.
bold L bold apostrophe bold Hospital bold apostrophe bold s bold space bold Rule bold space bold tells bold space bold us bold space bold that bold space bold if bold space bold we bold space bold have bold space bold an bold space bold indeterminate bold space bold form bold space bold 0 over bold 0 bold space bold all bold space bold we bold space bold need bold space bold to bold space bold do bold space bold is bold space
bold differentiate bold space bold the bold space bold numerator bold space bold and bold space bold differentiate bold space bold the bold space bold denominator bold space bold and bold space bold then bold space bold take bold space bold the bold space bold limit bold.
So comma space limit as straight x rightwards arrow 0 of fraction numerator square root of 1 plus straight x end root minus 1 over denominator cube root of 1 plus straight x end root minus 1 end fraction
equals limit as straight x rightwards arrow 0 of fraction numerator begin display style straight d over dx open parentheses square root of 1 plus straight x end root minus 1 close parentheses end style over denominator straight d over dx open parentheses cube root of 1 plus straight x end root minus 1 close parentheses end fraction
equals limit as straight x rightwards arrow 0 of fraction numerator begin display style fraction numerator 1 over denominator 2 square root of 1 plus straight x end root end fraction straight d over dx left parenthesis 1 plus straight x right parenthesis minus 0 end style over denominator 1 third open parentheses 1 plus straight x close parentheses to the power of begin display style 1 third minus 1 end style end exponent straight d over dx left parenthesis 1 plus straight x right parenthesis minus 0 end fraction
equals limit as straight x rightwards arrow 0 of fraction numerator begin display style fraction numerator 1 over denominator 2 open parentheses 1 plus straight x close parentheses to the power of 1 half end exponent end fraction end style over denominator 1 third open parentheses 1 plus straight x close parentheses to the power of begin display style fraction numerator negative 2 over denominator 3 end fraction end style end exponent end fraction equals limit as straight x rightwards arrow 0 of fraction numerator begin display style 1 half end style over denominator 1 third end fraction fraction numerator 1 over open parentheses 1 plus straight x close parentheses to the power of 1 half end exponent over denominator begin display style 1 over open parentheses 1 plus straight x close parentheses to the power of fraction numerator negative 2 over denominator 3 end fraction end exponent end style end fraction equals limit as straight x rightwards arrow 0 of fraction numerator begin display style 3 end style over denominator 2 end fraction 1 over open parentheses 1 plus straight x close parentheses to the power of 1 half minus 2 over 3 end exponent
equals limit as straight x rightwards arrow 0 of fraction numerator begin display style 3 end style over denominator 2 end fraction 1 over open parentheses 1 plus straight x close parentheses to the power of fraction numerator negative 1 over denominator 6 end fraction end exponent equals fraction numerator begin display style 3 end style over denominator 2 end fraction end style
 
Hi Student, the answer is 3/2. Kindly check. Thank you.
Answered by Expert 14th March 2017, 9:45 PM
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