\[ \lim_{x \to 0} 2 \cdot \frac{\left( 1 - \frac{x^2}{2!} \right) \left( 1 - \frac{4x^2}{2!} \right) \left( 1 - \frac{9x^2}{2!} \right) \cdots \left( 1 - \frac{100x^2}{2!} \right)}{x^2} \]
By expansion:
\[ \lim_{x \to 0} 2 \cdot \frac{\left[ 1 - \frac{x^2}{2} \right] \left[ 1 - \frac{2x^2}{2} \right] \left[ 1 - \frac{3x^2}{2} \right] \cdots \left[ 1 - \frac{10x^2}{2} \right]}{x^2}. \]
Simplify the product:
\[ \lim_{x \to 0} 2 \cdot \frac{1 - \left[ \frac{x^2}{2} + \frac{2x^2}{2} + \frac{3x^2}{2} + \cdots + \frac{10x^2}{2} \right]}{x^2}. \]
The \(x^2\) terms cancel:
\[ 2 \cdot \left( \frac{1}{2} + \frac{2}{2} + \frac{3}{2} + \cdots + \frac{10}{2} \right). \]
Simplify the summation:
\[ 2 \cdot \frac{1 + 2 + 3 + \cdots + 10}{2}. \]
The sum of the first 10 natural numbers is:
\[ \frac{10 \cdot 11}{2} = 55. \]
Final Answer: 55.
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:
If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
The magnitude of heat exchanged by a system for the given cyclic process ABC (as shown in the figure) is (in SI units):
