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Sir pls solve the following.
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Asked by rsudipto | 04 Jan, 2019, 19:16: PM
answered-by-expert Expert Answer
begin mathsize 12px style limit as x rightwards arrow 0 of space fraction numerator ln left parenthesis x plus a right parenthesis minus ln left parenthesis a right parenthesis over denominator x end fraction space equals space limit as x rightwards arrow 0 of fraction numerator ln begin display style fraction numerator x plus a over denominator a end fraction end style over denominator x end fraction space equals space limit as x rightwards arrow 0 of fraction numerator ln open parentheses 1 plus begin display style x over a end style close parentheses over denominator x end fraction space space equals space 1 over a end style
begin mathsize 12px style L e t space x minus e space equals space y limit as x rightwards arrow e of fraction numerator ln space x space minus 1 over denominator x minus e end fraction space equals space limit as y rightwards arrow 0 of fraction numerator ln left parenthesis y plus e right parenthesis minus ln e over denominator y end fraction space equals space limit as y rightwards arrow 0 of fraction numerator ln open parentheses 1 plus begin display style y over e end style close parentheses over denominator y end fraction space equals space 1 over e end style
 
begin mathsize 12px style limit as x rightwards arrow 0 of space fraction numerator ln left parenthesis x plus a right parenthesis minus ln left parenthesis a right parenthesis over denominator x end fraction space end style + k begin mathsize 12px style limit as x rightwards arrow e of fraction numerator ln space x space minus 1 over denominator x minus e end fraction space space equals space 1 over a space plus space k over e space equals space 1 end style
 
hence k = begin mathsize 12px style e space open parentheses 1 space minus space 1 over a close parentheses end style
Answered by Thiyagarajan K | 07 Jan, 2019, 15:58: PM

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