Question #120b5

1 Answer
Jun 4, 2016

Answer:

#lim_{x->infty}31/5 x (Log_e(x + 1) - Log_e(x)) = 31/5#

Explanation:

Calling #f(x) = 31/5 x (Log_e(x + 1) - Log_e(x))# and performing simplifications

#f(x) = log_e((x+1)/x)^{31/5 x} = log_e(((x+1)/x)^x)^{31/5}#

then

#lim_{x->infty}f(x) = log_e(lim_{x->infty}((x+1)/x)^x)^{31/5}#

but (this is a typical textbook result)

#lim_{x->infty}((x+1)/x)^x = lim_{x->infty}(1+1/x)^x = e#

finally

#lim_{x->infty}f(x) =log_e(e^{31/5}) = 31/5#