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
