We know that: #Lim_(xrarroo)(1+1/x)^x=e#
#Lim_(xrarroo)((x^2+4x+1)/(x^2+2x+2))^(10x)#
#Lim_(xrarroo)((x^2+4x+1+2x-2x+1-1)/(x^2+2x+2))^(10x)#
#Lim_(xrarroo)((x^2+2x+2)/(x^2+2x+2)+(2x-1)/(x^2+2x+2))^(10x)#
#Lim_(xrarroo)(1+1/((x^2+2x+2)/(2x-1)))^(10x)#
#Lim_(xrarroo)(1+1/((x^2+2x+2)/(2x-1)))^((10x)((x^2+2x+2)/(2x-1))((2x-1)/(x^2+2x+2))#
Looks very messy but I like it this way. You can substitute to make it more clear but as you can see now we have the same formula as in the beginning.
#Lim_(xrarroo)[(1+1/((x^2+2x+2)/(2x-1)))^((x^2+2x+2)/(2x-1))]^((10x)((2x-1)/(x^2+2x+2))#
We know that: #Lim_(xrarroo)[(1+1/((x^2+2x+2)/(2x-1)))^((x^2+2x+2)/(2x-1))]=e#
And now we can put limit to the exponent (superscript)
#e^(Lim_(xrarroo)(10x)((2x-1)/(x^2+2x+2)))=?#
Let's evaluate limit and then finish it
#Lim_(xrarroo)(10x)((2x-1)/(x^2+2x+2))#
#Lim_(xrarroo)((20x^2-10x)/(x^2+2x+2))#
#Lim_(xrarroo)(cancel(x^2)(20-10/x))/(cancel(x^2)(1+2/x+2/x^2))#
#(20-10/oo)/(1+2/oo+2/oo^2)=(20-0)/(1+0+0)=20/1=20#
The answer: #Lim_(xrarroo)((x^2+4x+1)/(x^2+2x+2))^(10x)=e^20#