solve the limit using the binomial theorem: lim x --> 0 (5x-2)^3+8 / (2x+3)^3 -27

Question

 solve the limit using the binomial theorem:

 

lim        (5x-2)^3+8 / (2x+3)^3 -27

x --> 0

Answer 1

 

 

(A+B)^3 = (3 choose 0) A^3 B^0 + (3 choose 1) A^2 B + (3 choose 2) A B^2 + (3 choose 3) B^3

 

(3 choose 0) = (3 choose 3) = N^0 = 1 where N is any non-zero real number

 

(3 choose 1) = (3 choose 2) = 3 by property of combinatorics and factorials 

 

The expansion simplifies to A^3 + 3 A^2 B + 3 A B^2 + B^3

 

which is further supported by Pascal's Triangle

 

                      1

                   1     1

                 1    2    1    <--- squared

              1   3     3    1  <--- cubed

 

The powers of A begin at three and decrease while

the powers of B begin at zero and increase.

 

Substituting A = 5x and B = -2 for the polynomial in the numerator

results in the expansion

 

125x^3 - 150x^2 + 60x - 8

 

Substituting A=2x and B=3 for the polynomial in the denomiantor

results in the expansion 

 

8x^3 + 36x^2 +54x + 27

 

Brute force algebra, FOIL method, and polynomial multiplication

support these results.

 

BUT, the constant terms cancel when 8 is added in the numerator

and 27 is subtracted in the denominator after plugging these polynomials in.

 

So x can be factored out of both polynomials, which is reduced to:

  

   125x^2 + 150x  + 60

----------------------------

    8x^2 + 36x + 54

 

As x approaches zero, the limit is 60/54 = 10/9

 

-------------------------------------------------------------

 

As a check, L'Hopital's rule is applied to the original function:

 

THe derivative of the numerator is 3(5x-2)^2*5  by chain rule

The derivative of the bottom is 3(2x+3)^2 * 2  by chain rule

 

As x approaches zero, the limit is   3(-2)^2*5  divided by 3(3)^2*2 

 

                                         =       3*4*5  / 3*9*2

 

                                        =  60/54

 

                                         = 10/9