If _ x → ∞ a e ^ x + bcosx + c + dx xsi n ^ 2 x = 3 , then the...

Using expansions, we get,
$\lim_{x \to 0}\dfrac{a\left(1 + x + \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{3}}{3 !} + . .\right) + b\left(1 - \dfrac{{x}^{2}}{2 !} + . . .\right) + c + dx}{x{\left(x - \dfrac{{x}^{3}}{3 !} + . . . . .\right)}^{2}} = 3$
$\lim_{x \to 0}\dfrac{\left(a + b + c\right) + \left(a + d\right)x + \left(\dfrac{a - b}{2}\right){x}^{2} + \dfrac{a}{6}{x}^{3} + \ldots}{{x}^{3}{\left(1 - \dfrac{{x}^{2}}{3 !} + . . . . .\right)}^{2}} = 3$
$∵$ in the denominator lowest power of $x$ is $3$
For the limit to be finite, the numerator should also have the least power of $x$ as $3$
$∴ a + b + c = 0 . . . \left(1\right)$
$a + d = 0 . . . \left(2\right)$
$\dfrac{a - b}{2} = 0 . . . \left(3\right)$
Now, $\dfrac{\left(\dfrac{a}{6}\right)}{1} = 3 ⇒ a = 18$
From \left(1\right), \left(2\right), \left(3\right)$, we get,</p><p>$a = 18, b = 18, c = – 36, d = – 18$<br />$\dfrac{abd}{{c}^{3}} = \dfrac{- {\left(18\right)}^{3}}{- 8{\left(18\right)}^{3}} = \dfrac{1}{8}