Let L = stack l i m with blank below x rightwards arrow space 0 (a - square root of a ² space minus space x ² end root - x²∕4) ÷ (x² × x²) , a > 0 . If L is finite , then (a) a = 2 , (b) a = 1 , (c) L = 1∕64 , (d)L = 1/32

Asked by yeeshuraj3 | 28th Jul, 2016, 03:46: PM

Expert Answer:

limit as straight x rightwards arrow 0 of fraction numerator open parentheses a minus square root of a squared minus x squared end root minus begin display style x squared over 4 end style close parentheses over denominator x to the power of 4 end fraction equals stack l i m with straight x rightwards arrow 0 below fraction numerator open parentheses 4 a minus x squared minus 4 square root of a squared minus x squared end root close parentheses over denominator 4 x to the power of 4 end fraction

Rationalize space the space numerator
equals limit as straight x rightwards arrow 0 of fraction numerator open parentheses negative 8 straight a plus straight x squared plus 16 close parentheses over denominator 4 straight x squared open parentheses 4 straight a minus straight x squared plus 4 square root of straight a squared minus straight x squared end root close parentheses end fraction
equals 1 fourth limit as straight x rightwards arrow 0 of 1 over straight x squared limit as straight x rightwards arrow 0 of fraction numerator open parentheses negative 8 straight a plus straight x squared plus 16 close parentheses over denominator open parentheses 4 straight a minus straight x squared plus 4 square root of straight a squared minus straight x squared end root close parentheses end fraction
equals 1 fourth open square brackets fraction numerator limit as straight x rightwards arrow 0 of open parentheses negative 8 straight a plus straight x squared plus 16 close parentheses limit as straight x rightwards arrow 0 of 1 over straight x squared over denominator limit as straight x rightwards arrow 0 of open parentheses 4 straight a minus straight x squared plus 4 square root of straight a squared minus straight x squared end root close parentheses end fraction close square brackets
equals 1 fourth open square brackets fraction numerator open parentheses negative 8 straight a plus 16 close parentheses limit as straight x rightwards arrow 0 of 1 over straight x squared over denominator open parentheses 4 straight a plus 4 square root of straight a squared end root close parentheses end fraction close square brackets

But space limit as straight x rightwards arrow 0 of 1 over straight x squared equals infinity

therefore limit as straight x rightwards arrow 0 of fraction numerator open parentheses straight a minus square root of straight a squared minus straight x squared end root minus begin display style straight x squared over 4 end style close parentheses over denominator straight x to the power of 4 end fraction equals fraction numerator infinity open parentheses negative 8 straight a plus 16 close parentheses over denominator 4 open parentheses 4 straight a plus 4 square root of straight a squared end root close parentheses end fraction

Here space we space are space getting space value space of space straight L space as space space infinite space term.

If space we space take space straight L space as space finite space equals space fraction numerator open parentheses negative 8 straight a plus 16 close parentheses over denominator 4 open parentheses 4 straight a plus 4 square root of straight a squared end root close parentheses end fraction space then space also space it space is space not space satisfing space given space options.

Please space check space the space options.

Answered by Vijaykumar Wani | 29th Jul, 2016, 01:07: PM