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Please help to find k given in the problem
 
Asked by rajuharan 3rd July 2016, 10:14 AM
Answered by Expert

Answer:
limit as straight x rightwards arrow 0 to the power of plus of straight f left parenthesis straight x right parenthesis space equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator square root of 1 plus kx end root space minus space square root of 1 minus kx end root over denominator straight x end fraction
equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator open square brackets square root of 1 plus kx end root space minus space square root of 1 minus kx end root close square brackets open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction
equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator open square brackets 1 plus kx minus 1 plus kx close square brackets over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction
equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator 2 kx over denominator straight x open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction
equals limit as straight x rightwards arrow 0 to the power of plus of fraction numerator 2 straight k over denominator open square brackets square root of 1 plus kx end root space plus space square root of 1 minus kx end root close square brackets end fraction
equals fraction numerator 2 straight k over denominator open square brackets square root of 1 plus 0 end root space plus space square root of 1 minus 0 end root close square brackets end fraction
equals straight k


space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis equals space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis fraction numerator 2 straight x plus 1 over denominator straight x minus 2 end fraction equals fraction numerator 0 plus 1 over denominator 0 minus 2 end fraction equals negative 1 half

Given space that space straight f left parenthesis straight x right parenthesis space is space continuous space at space space straight x space equals space 0

therefore limit as straight x rightwards arrow 0 to the power of plus of straight f left parenthesis straight x right parenthesis space space equals space limit as straight x rightwards arrow 0 to the power of minus of straight f left parenthesis straight x right parenthesis space
rightwards double arrow straight k space equals space minus 1 half
Answered by Expert 4th July 2016, 10:50 AM
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