Determine if L'Hopital's Rule can be used. If it can, solve the limit. If it cannot choose the appropriate response. \( \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x} \) -1 0 L'Hopital's Rule does not apply. 2

To determine if L'Hopital's Rule can be used, we first check the limit along with the form of the functions involved. As \( x \) approaches 0, both the numerator \( e^{x} - e^{-x} \) and the denominator \( \sin x \) approach 0, thus yielding the indeterminate form \( \frac{0}{0} \), allowing us to apply L'Hopital's Rule. Now, applying L'Hopital's Rule involves differentiating the numerator and denominator. The derivative of the numerator, \( e^x + e^{-x} \), simplifies to \( \cosh(x) \), and the derivative of the denominator, \( \cos x \), leads to: \[ \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \frac{e^0 + e^0}{\cos(0)} = \frac{1 + 1}{1} = 2. \] So, the final answer is **2**.