Calculus: Early Transcendentals, Chapter 4, 4.4, Section 4.4, Problem 66

Replacing oo for x in limit equation yields the nedetermination oo^oo . You need to use the following technique, such that:
f(x) = ((2x - 3)/(2x + 5))^(2x + 1)
lim_(x->oo) ((2x - 3)/(2x + 5))^(2x + 1) = lim_(x->oo) ((2x + 5 - 5 - 3)/(2x + 5))^(2x + 1)
lim_(x->oo) ((2x + 5)/(2x + 5) - 8/(2x + 5))^(2x + 1)
lim_(x->oo) (1 + (- 8)/(2x + 5))^(2x + 1)
Since lim_(x->oo) (- 8)/(2x + 5) = -8/oo = 0, yields:
lim_(x->oo) (1 + (- 8)/(2x + 5))^(2x + 1) = 1^oo
You need to remember that lim_(x->oo) (1 + 1/x)^x = e , hence, you need to re-create the special limit:
lim_(x->oo) ((1 + (-8)/(2x+5))^((2x+5)/(-8)))^(-8(2x+1)/(2x+5))
You need to notice that lim_(x->oo)(1+1/x)^x = oo , such that:
lim_(x->oo)((1 + (-8)/(2x+5))^(((2x+5)/(-8))))^(-8(2x+1)/(2x+5)) = e^lim_(x->oo) (-8(2x+1)/(2x+5))
You need to evaluate lim_(x->oo)(- 8)*(2x+1)/(2x + 5) :
lim_(x->oo)(- 8)*(2x+1)/(2x + 5) = -8*lim_(x->oo)(2x+1)/(2x + 5) = -8*(2/2) = -8
e^lim_(x->oo)(- 8)*(2x+1)/(2x + 5) = e^(-8)
Hence, evaluating the given limit, using special limit, yields lim_(x->oo) ((2x - 3)/(2x + 5))^(2x + 1) = 1/(e^8).