If $lim x \to \infty \frac{a (2 x^{3} - x^{2}) + b (x^{3} - 1) - c (3 x^{3} + x^{2})}{a (5 x^{4} - x) - b x^{4} + c (4 x^{4} + 1) + 2 x^{2} + 5 x} = 1,$ then the value of $(a - b - c)$ can be expressed in the lowest form as $\frac{p}{q}$ where $p, q \in N .$ The value of $p + q$ is

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Solution

$lim x \to \infty \frac{a (2 x^{3} - x^{2}) + b (x^{3} - 1) - c (3 x^{3} + x^{2})}{a (5 x^{4} - x) - b x^{4} + c (4 x^{4} + 1) + 2 x^{2} + 5 x} = 1$

$lim x \to \infty \frac{(2 a + b - 3 c) x^{3} + (- a - c) x^{2} - b}{(5 a - b + 4 c) x^{4} + 2 x^{2} + (- a + 5) x + c} = 1$

For limit to exist and equal to $1,$
coefficient of $x^{4}$ in denominator $= 0$
$\Rightarrow 5 a - b + 4 c = 0 \dots (1)$
and coefficient of $x^{3}$ in numerator $= 0$
$\Rightarrow 2 a + b - 3 c = 0 \dots (2)$
and $\frac{coefficient of x^{2} in numerator}{coefficient of x^{2} in denominator} = 1$
$\Rightarrow \frac{- a - c}{2} = 1$
$\Rightarrow - a - c = 2 \dots (3)$

Solving $(1), (2)$ and $(3),$ we get
$a = \frac{1}{3}, b = - \frac{23}{3}$ and $c = - \frac{7}{3}$
$\Rightarrow a - b - c = \frac{31}{3} = \frac{p}{q}$
$∴ p + q = 34$

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