Asked by Topperlearning User | 10th Nov, 2016, 03:14: AM

Expert Answer:

limit as x rightwards arrow 1 of open parentheses log subscript 5 space 5 x close parentheses to the power of log subscript x 5 end exponent
equals limit as x rightwards arrow 1 of open parentheses log subscript 5 5 plus log subscript 5 x close parentheses to the power of log subscript x 5 end exponent
equals limit as x rightwards arrow 1 of open parentheses 1 plus log subscript 5 x close parentheses to the power of log subscript x 5 end exponent
equals limit as x rightwards arrow 1 of open parentheses 1 plus log subscript 5 x close parentheses to the power of fraction numerator 1 over denominator log subscript 5 x end fraction end exponent
I f space x rightwards arrow 1 comma space t h e n space log subscript 5 x rightwards arrow 0
L e t space log subscript 5 x equals y
equals limit as y rightwards arrow 0 of open parentheses 1 plus y close parentheses to the power of 1 over y end exponent
equals e to the power of limit as y rightwards arrow 0 of 1 over y log left parenthesis 1 plus y right parenthesis end exponent
to the power of limit as y rightwards arrow 0 of 1 over y log left parenthesis 1 plus y right parenthesis end exponent equals 1 space s o
e to the power of limit as y rightwards arrow 0 of 1 over y log left parenthesis 1 plus y right parenthesis end exponent equals e

Answered by  | 10th Nov, 2016, 05:14: AM