1. Simplify the numerator (2x2−ax+1)−(ax2+3bx+2)=(2−a)x2−(a+3b)x−1. So for x≠−1, f(x)=‌
(2−a)x2−(a+3b)x−1
x+1
. 2. Ensure the limit at x=−1 is finite For continuity, the limit as x⟶−1 must exist and be finite. Since the denominator is x+1, the numerator must vanish at x=−1 : (2−a)(−1)2−(a+3b)(−1)−1=0. Compute: (2−a)+(a+3b)−1=0⇒1+3b=0⇒b=−‌
1
3
. 3. Factor and find the limit With b=−‌
1
3
, the numerator is (2−a)x2−(a−1)x−1. We know x=−1 is a root, so (2−a)x2−(a−1)x−1=(x+1)((2−a)x−1). Thus, for x≠−1, f(x)=(2−a)x−1. So
lim
x⟶−1
f(x)=(2−a)(−1)−1=a−3. 4. Continuity at x=−1 For continuity: k=