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Continuity And Differentiability

Question
CBSEENMA12034494
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The function straight f left parenthesis straight x right parenthesis equals fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fractionis not defined at x = 0. Find the value of/(x) so that f is continuous at x = 0.

Solution
Here space straight f left parenthesis straight x right parenthesis equals fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 below fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator log left parenthesis 1 plus ax right parenthesis over denominator straight x end fraction minus bold Lt with bold x bold rightwards arrow bold 0 below fraction numerator bold log bold left parenthesis bold 1 bold minus bold bx bold right parenthesis over denominator bold x end fraction space space space space space space space space space space space space space equals straight a space Lt with straight x rightwards arrow 0 below 1 over ax log left parenthesis 1 plus ax right parenthesis minus straight b. Lt with straight x rightwards arrow 0 below 1 over bx log left parenthesis 1 minus bx right parenthesis space space space space space space space space space space space space space equals straight a space Lt with straight x rightwards arrow 0 below log left parenthesis 1 plus ax right parenthesis to the power of 1 over ax end exponent plus straight b. Lt with straight x rightwards arrow 0 below log left parenthesis 1 minus bx right parenthesis to the power of negative 1 over bx end exponent space space space space space space space space space space space space space equals straight a space log space straight e plus straight b space log space straight e equals straight a plus straight b
For f(x) to be continuous at x = 0, we have
straight f left parenthesis 0 right parenthesis equals Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space space space space space space space straight i. straight e. comma space straight f left parenthesis 0 right parenthesis equals straight a plus straight b therefore space function space must space be space defined space as space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator log left parenthesis 1 plus ax right parenthesis minus log left parenthesis 1 minus bx right parenthesis over denominator straight x end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space space space space space space straight a plus straight b comma space space space space space space space space space space space space space space space straight x equals 0 end cell end table close

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