Final answer:
The limit of \(e^y \cdot \frac{\sin(2x)}{x}\) as \(x\) and \(y\) approach 0 is 1. This is found by separately evaluating the limits of \(e^y\) and \(\frac{\sin(2x)}{x}\) as the variables approach 0. The correct option is A.
Step-by-step explanation:
The student is asking to evaluate the limit of the function \(e^y \cdot \frac{\sin(2x)}{x}\) as both \(x\) and \(y\) approach 0. To find this limit, we first recognize that the \(e^y\) term approaches 1 as \(y\) goes to 0, because the exponent of \(e\) is approaching 0, and \(e^0 = 1\).
Next, we focus on the \(\frac{\sin(2x)}{x}\) term. As \(x\) approaches 0, this approaches the well-known limit \(\lim_{x\to0}\frac{\sin(x)}{x} = 1\), except with \(2x\) instead of \(x\), which still yields the same result since \(2x\) approaches 0 when \(x\) approaches 0.
Applying these findings to our original function, \(e^y \cdot \frac{\sin(2x)}{x}\) becomes \(1 \cdot 1 = 1\) as both \(x\) and \(y\) approach 0. Hence, the answer to the limit is b) 1.
It is important to note that the provided reference information seems unrelated to the calculation of this limit and appears to discuss quantum mechanics and wave functions, which are not directly applicable to this limit problem.