If lim _{x rightarrow 0} frac{ae^{x}-b cos x | vedclass

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(D) Given $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2.$Since the denominator $x \sin x \approx x^2$ as $x \rightarrow 0$,t

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$4$

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(D) Given $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2.$Since the denominator $x \sin x \approx x^2$ as $x \rightarrow 0$,the numerator must also approach $0$ at a rate of at least $x^2$ for the limit to exist.Using Taylor series expansions:$e^x = 1 + x + \frac{x^2}{2} + \dots$$\cos x = 1 - \frac{x^2}{2} + \dots$$e^{-x} = 1 - x + \frac{x^2}{2} + \dots$Substituting these into the numerator:$a(1 + x + \frac{x^2}{2}) - b(1 - \frac{x^2}{2}) + c(1 - x + \frac{x^2}{2}) = (a - b + c) + (a - c)x + (\frac{a+b+c}{2})x^2 + \dots$For the limit to be finite,the coefficients of $x^0$ and $x^1$ must be $0$:$a - b + c = 0$ and $a - c = 0 \Rightarrow a = c$.Substituting $a = c$ into the first equation: $a - b + a = 0 \Rightarrow b = 2a$.The limit becomes $\lim _{x \rightarrow 0} \frac{(\frac{a+b+c}{2})x^2}{x^2} = \frac{a+b+c}{2} = 2$.Thus,$a + b + c = 4$.

If $\lim _{x \rightarrow 0} \frac{ae^{x}-b \cos x + ce^{-x}}{x \sin x} = 2,$ then $a + b + c$ is equal to ...........

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