How do you evaluate #((2x-1)/(2x+5))^(2x+3)# as x approaches infinity?

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Answer 1 · 2016-06-07T11:01:11

#lim_{x->infty}( (2x-1)/(2x+5))^{2x+3} = e^{-6}#

Making #y = 2x-1# in #f(x) =( (2x-1)/(2x+5))^{2x+3}# we get

#g(y) = (y/(y+6))^{y+4} = (y/(y+6))^y (y/(y+6))^4 #

but

# (y/(y+6))^y = (1/(1+6/y))^y = ( (1/(1+6/y))^{y/6})^6#

then

#lim_{x->infty}f(x) equiv lim_{y->infty}g(y)#

and

#lim_{y->infty}g(y) = (lim_{y->infty} (1/(1+6/y))^{y/6})^6(lim_{y->infty}( 1/(1+6/y)))^4 #

here
#lim_{y->infty}( 1/(1+6/y)) =1#

From any Calculus textbook we get

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

so

#lim_{y->infty}g(y) = e^{-6}#