The problem is:
#lim_(x->oo)((2x+2)/(2x+4))^(2x+2)#
Now this is a difficult problem. The solution lies in very careful pattern recognition. You may recall the definition of #e#:
#e=lim_(u->oo)(1+1/u)^u~~2.718...#
If we could rewrite the limit as something close to the definition of #e#, we would have our answer. So, let's try it.
Note that #lim_(x->oo)((2x+2)/(2x+4))^(2x+2)# is equivalent to:
#lim_(x->oo)((2x+4-2)/(2x+4))^(2x+2)#
We can split up the fractions like so:
#lim_(x->oo)((2x+4)/(2x+4)-2/(2x+4))^(2x+2)#
#=lim_(x->oo)(1-2/(2x+4))^(2x+2)#
We're getting there! Let's factor out a #-2# from the top and bottom:
#lim_(x->oo)(1-2/(2x+4))^(2x+2)#
#=lim_(x->oo)(1+((-2))/(-2(-x-2)))^(2x+2)#
#->lim_(x->oo)(1+(cancel(-2))/(cancel(-2)(-x-2)))^(2x+2)#
#=lim_(x->oo)(1+1/(-x-2))^(2x+2)#
Let us apply the substitution #u=-x-2->x=-2-u#:
#lim_(x->oo)(1+1/(-x-2))^(2x+2)#
#=(1+1/u)^(2(-2-u)+2#
#=(1+1/u)^(-4-2u+2)#
#=(1+1/u)^(-2u-2)#
The properties of exponents say: #x^(a+b)=x^ax^b#
So #lim_(x->oo)(1+1/u)^(-2u-2)# is equivalent to:
#lim_(x->oo)(1+1/u)^(-2u)(1+1/u)^(-2)#
The properties of exponents also say that: #x^(ab)=x^(a^b)#
Which means this further reduces to:
#lim_(x->oo)(1+1/u)^((u)^(-2))(1+1/u)^(-2)#
#=lim_(x->oo)(1+1/u)^((u)^(-2))lim_(x->oo)(1+1/u)^(-2)#
By definition, #lim_(x->oo)(1+1/u)^(u)=e#; and using direct substitution on the second limit yields:
#lim_(x->oo)(1+1/u)^(-2)#
#=1/(1+1/oo)^(2)#
#=1/(1+0)^(2)#
#=1/1^(2)=1#
So the solution is...
#lim_(x->oo)(1+1/u)^((u)^(-2))lim_(x->oo)(1+1/u)^(-2)#
#=(e)^-2(1)#
#=e^-2#
