Question #0df97

1 Answer
Jul 17, 2016

Answer:

The answer to 4 is #e^-2#.

Explanation:

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#