If displaystyle lim x arrow ∞((x3+1/x2+1)-(a x +b))=2, then find the value of 10 a+3 b.

Question

If displaystyle lim x arrow ∞((x3+1/x2+1)-(a x +b))=2, then find the value of 10 a+3 b. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

displaystyle lim x arrow ∞((x3+1/x2+1)-a x-b) = displaystyle lim x arrow ∞((x3(1-a)-b x2-a x+1-b/x2+1)) Given that the limit is finite ⇒ 1-a=0 and b=2 ⇒ a=1 and b=-2 ⇒ 10 a+3 b=10-6=4

Q. If $x \to \infty lim (\frac{x ^{3} + 1}{x ^{2} + 1} - (a x + b)) = 2$ , then find the value of $10 a + 3 b$ .

$x \to \infty lim (\frac{x ^{3} + 1}{x ^{2} + 1} - a x - b)$
$= x \to \infty lim (\frac{x ^{3} ( 1 - a ) - b x ^{2} - a x + 1 - b}{x ^{2} + 1})$
Given that the limit is finite
$\Rightarrow 1 - a = 0$ and $b = 2$
$\Rightarrow a = 1$ and $b = - 2$
$\Rightarrow 10 a + 3 b = 10 - 6 = 4$

Q. If x→∞lim​(x2+1x3+1​−(ax+b))=2, then find the value of 10a+3b.

x→∞lim​(x2+1x3+1​−ax−b) =x→∞lim​(x2+1x3(1−a)−bx2−ax+1−b​) Given that the limit is finite ⇒1−a=0 and b=2 ⇒a=1 and b=−2 ⇒10a+3b=10−6=4

Q. If $x \to \infty lim (\frac{x ^{3} + 1}{x ^{2} + 1} - (a x + b)) = 2$ , then find the value of $10 a + 3 b$ .

$x \to \infty lim (\frac{x ^{3} + 1}{x ^{2} + 1} - a x - b)$
$= x \to \infty lim (\frac{x ^{3} ( 1 - a ) - b x ^{2} - a x + 1 - b}{x ^{2} + 1})$
Given that the limit is finite
$\Rightarrow 1 - a = 0$ and $b = 2$
$\Rightarrow a = 1$ and $b = - 2$
$\Rightarrow 10 a + 3 b = 10 - 6 = 4$