\begin{tabular}{l} 2, 4.5-4.9) password: \\ \( \qquad \) Evaluate the following limit. Use l'Hôpital's Rule when it is convenient and applicabl \\ \( \lim _{x \rightarrow 0} \frac{1-\cos 7 x}{6 x^{2}} \) \\ \hline\end{tabular}

To evaluate the limit \( \lim_{x \rightarrow 0} \frac{1 - \cos 7x}{6x^2} \), we start by noticing that as \( x \) approaches 0, both the numerator and the denominator approach 0. This gives us an indeterminate form of \( \frac{0}{0} \), which means we can apply l'Hôpital's Rule. Applying l'Hôpital's Rule, we differentiate the numerator and the denominator: 1. The derivative of the numerator \( 1 - \cos 7x \) is \( 7\sin 7x \). 2. The derivative of the denominator \( 6x^2 \) is \( 12x \). So, our limit now looks like this: \[ \lim_{x \rightarrow 0} \frac{7\sin 7x}{12x} \] As \( x \) approaches 0, we plug in values: \[ \lim_{x \rightarrow 0} \frac{7\sin 7x}{12x} = \lim_{x \rightarrow 0} \frac{7 \cdot 7\cos 7x}{12} = \frac{7 \cdot 7 \cdot 1}{12} = \frac{49}{12} \] Thus, the final result for the limit is: \[ \lim_{x \rightarrow 0} \frac{1-\cos 7x}{6x^2} = \frac{49}{12} \]