If _ x → 0 a e ^ x - bcosx + c e ^ - x xsinx = 2 , then a + b + c...

Given $\lim_{x \to 0}\dfrac{a{e}^{x} - bcosx + c{e}^{- x}}{xsinx} = 2$

$⇒ \lim_{x \to 0}\dfrac{a\left(1 + x + \dfrac{{x}^{2}}{2 !} \ldots\right) - b\left(1 - \dfrac{{x}^{2}}{2 !} + \ldots\right) + c\left(1 - x + \dfrac{{x}^{2}}{2 !}\right)}{\left(\dfrac{xsinx}{x}\right)x} = 2$

$⇒ \lim_{x \to 0}\dfrac{a\left(1 + x + \dfrac{{x}^{2}}{2 !} \ldots\right) - b\left(1 - \dfrac{{x}^{2}}{2 !} + \ldots\right) + c\left(1 - x + \dfrac{{x}^{2}}{2 !}\right)}{{x}^{2}} = 2\left(∵ \lim_{x \to 0}\dfrac{\sin x}{x} = 1\right)$

For the given limit to exist numerator should be $0, \text{when}x \to 0 .$

Then, $a - b + c = 0$ $. . . ( i )$

Applying L'Hospital's rule, we get

$\lim_{x \to 0}\dfrac{a\left(1 + \dfrac{2x}{2 !} \ldots\right) - b\left(- \dfrac{2x}{2 !} + \ldots\right) + c\left(- 1 + \dfrac{2x}{2 !}\right)}{2x} = 2$

Again, for the given limit to exist numerator should be $0, \text{when}x \to 0 .$

Then, $a - c = 0 . . . \left(ii\right)$

Applying, L'Hospital's rule, we get

$\lim_{x \to 0}\dfrac{a\left(1 + \ldots\right) - b\left(- 1 + \ldots\right) + c\left(1\right)}{2} = 2$

$⇒ \dfrac{a + b + c}{2} = 2$

$⇒ a + b + c = 4$