Question:
If \(\lim\limits_{x \to 0} \frac{a e^x - b\cos x + c e^{-x}}{x \sin x} = 2\), then \(a + b + c\) is equal to ____________.
If \(\lim\limits_{x \to 0} \frac{a e^x - b\cos x + c e^{-x}}{x \sin x} = 2\), then \(a + b + c\) is equal to ____________.
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The denominator behaves like \(x^2\); the numerator's constant and x terms must vanish, and the \(x^2\) coefficient must equal 2.
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
Expand: \(ae^x - b\cos x + ce^{-x} = (a - b + c) + (a - c)x + \frac{a + b + c}{2}x^2 + \dots\), and \(x\sin x \approx x^2\).
For a finite limit: \(a - b + c = 0\) and \(a - c = 0\).
Then the limit is \(\frac{a+b+c}{2} = 2\), so \(a + b + c = 4\).
(Consistent solution: a = c = 1, b = 2.)
Answer: 4
For a finite limit: \(a - b + c = 0\) and \(a - c = 0\).
Then the limit is \(\frac{a+b+c}{2} = 2\), so \(a + b + c = 4\).
(Consistent solution: a = c = 1, b = 2.)
Answer: 4
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