JEE PYQ: Limits Question 1
Question 1 - 2021 (16 Mar Shift 1)
If $\lim_{x \to 0} \frac{ae^x - b\cos x + ce^{-x}}{x \sin x} = 2$, then $a + b + c$ is equal to ______.
Show Answer
Answer: 4
Solution
Expanding using Taylor series: $ae^x = a\left(1 + x + \frac{x^2}{2} + \cdots\right)$, $b\cos x = b\left(1 - \frac{x^2}{2} + \cdots\right)$, $ce^{-x} = c\left(1 - x + \frac{x^2}{2} + \cdots\right)$. For limit to exist: $a - b + c = 0$ and $a - c = 0$. Then $\frac{a + b + c}{2} = 2 \Rightarrow a + b + c = 4$.
BETA
