To determine the values of $a$, $b$, and $c$ such that

$\lim_{x \to 0} \frac{a e^x - b \cos x + c e^{-x}}{x \sin x} = 2,$

we proceed as follows:

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Step 1: Simplify the given limit

Expand the denominator $x \sin x$ around $x = 0$ using the Taylor series: $\sin x \approx x - \frac{x^3}{6} \quad \Rightarrow \quad x \sin x \approx x^2 - \frac{x^4}{6}.$ For small $x$, we can approximate $x \sin x \approx x^2$, as higher-order terms become negligible.

For the numerator $a e^x - b \cos x + c e^{-x}$, use the Taylor expansions: $e^x \approx 1 + x + \frac{x^2}{2}, \quad \cos x \approx 1 - \frac{x^2}{2}, \quad e^{-x} \approx 1 - x + \frac{x^2}{2}.$ Substitute these expansions: $a e^x \approx a(1 + x + \frac{x^2}{2}), \quad b \cos x \approx b(1 - \frac{x^2}{2}), \quad c e^{-x} \approx c(1 - x + \frac{x^2}{2}).$ Combine these to get: $a e^x - b \cos x + c e^{-x} \approx (a + c - b) + (a - c)x + \frac{1}{2}(a + c - b)x^2.$

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Step 2: Substitute into the limit

The numerator becomes: $a e^x - b \cos x + c e^{-x} \approx (a + c - b) + (a - c)x + \frac{1}{2}(a + c - b)x^2.$ Dividing the numerator by $x^2$ (since $x \sin x \approx x^2$) gives: $\frac{a e^x - b \cos x + c e^{-x}}{x^2} \approx \frac{a + c - b}{x^2} + \frac{a - c}{x} + \frac{1}{2}(a + c - b).$

For the limit to exist, the coefficients of $\frac{1}{x^2}$ and $\frac{1}{x}$ must vanish:

1. $a + c - b = 0,$
2. $a - c = 0.$

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Step 3: Solve for $a$, $b$, and $c$

From $a - c = 0$, we get $a = c$.
From $a + c - b = 0$, substituting $c = a$, we have: $a + a - b = 0 \quad \Rightarrow \quad b = 2a.$

Now the limit becomes: $\lim_{x \to 0} \frac{a e^x - b \cos x + c e^{-x}}{x \sin x} = \frac{1}{2}(a + c - b) = \frac{1}{2}(a + a - 2a) = a.$

We are given that this equals 2, so $a = 2$. Hence: $a = 2, \quad b = 4, \quad c = 2.$

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Final Answer:

$a = 2, \quad b = 4, \quad c = 2.$

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Do you want details about the Taylor series expansion or simplifying steps? Let me know!

Here are 5 follow-up questions:

1. How are Taylor series used to approximate functions near a point?
2. What are higher-order terms in a Taylor expansion, and when can they be ignored?
3. Why is the denominator $x \sin x$ approximated as $x^2$ for small $x$?
4. How do we ensure limits exist by balancing terms in the numerator and denominator?
5. Can this method be applied to limits involving logarithmic or trigonometric terms?

Tip: Always start with Taylor expansions for small values of $x$ to simplify limit calculations near $x = 0$.