#(2x-3)^20 = 2^20x^20 +"terms of degree lass than 20"#
#(3x+2)^30 = 3^30x^30 + "terms of degree lass than 30"#
So the numerator, when we expand, will be
#2^20 3^30 x^50 + "terms of degree less than 50"#
and
#(2x+1)^50 = 2^50x^50 + "terms of degree less than 50"#
The ratio is:
#(2^20 3^30 x^50+ ("terms of degree less than 50"))/(2^50 x^50 + ("terms of degree less than 50"))#
Now, if we multiply by #(1/x^50)/(1/x^50)#, we get:
#(2^20 3^30 + ("terms of degree less than 50")/x^50)/(2^50 + "terms of degree less than 50"/x^50)#
As #xrarroo#, this goes to #(2^20 3^30)/2^50 = 3^30/2^30#
