In the calculation result of (AX ^ 2 + BX + 1) (3x ^ 2-2x + 1), there is no x ^ 3 term and X term, and the value of a and B is calculated

In the calculation result of (AX ^ 2 + BX + 1) (3x ^ 2-2x + 1), there is no x ^ 3 term and X term, and the value of a and B is calculated

(AX ^ 2 + BX + 1) (3x ^ 2-2x + 1) = 3ax ^ 4-2ax & # 179; + ax & # 178; + 3bx & # 179; - 2bx & # 178; + BX + 3x & # 178; - 2x + 1 = 3ax ^ 4 + (3b-2a) x & # 179; + (a-2b + 3) x & # 178; + (b-2) x + 1 because there is no x ^ 3 term and X term in the calculation result, 3b-2a = 0, ② substitute B = 2 from ② to get 3

Find the limit! Lim [x / √ (X & # 178; + 1)] why = 1?

lim[x/√(x²+1)]
=lim[1/√(1+1/x²)]
=1

Lim tends to infinity. (x + 2 / x + 3) ^ 2x

It should be [(x + 2) / (x + 3)] ^ 2x, right?
So this is the infinite power of 1
lim (x->oo) [1-1/(x+3)]^{-(x+3)*2x/[-(x+3)]}=e^lim (x->oo) (-2x)/(x+3)=e^(-2)