CBSE Class 11-science Answered
Let L = ![]() ![]() |
Asked by yeeshuraj3 | 28 Jul, 2016, 15:46: PM
![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.](https://images.topperlearning.com/topper/tinymce/cache/2457a1a2ea3437d0498547e59e5827ec.png)
Answered by Vijaykumar Wani | 29 Jul, 2016, 13:07: PM
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