Question

The values of constants 'a' and 'B so that lim
$( {x }^{2} - 1 \div x + 1 \: - ax - b ) = 2$
is​

Answer 1

lim(x+1x2+1−ax−b)=21

=x→∞lim(x+1(x2+1)−(ax+b)(x+1))=21

=x→∞lim(x+1(1−a)x2−(a+b)x−(b+1))=21

=x→∞lim(1+1/x(1−a)x−(a+b)−(b+1)/x2)=21

⇒a=1,a+b=−21⇒b=−23

Step-by-step explanation:

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