Find the following limit using \( \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 \). \( \lim _{x \rightarrow 0} \frac{\tan 5 x}{\sin 9 x} \) Select the correct choice below and, if necessary, fill in the answer box in your choice. A. \( \lim _{x \rightarrow 0} \frac{\tan 5 x}{\sin 9 x}=\square \) (Type an integer or a simplified fraction.) B. The limit does not exist.

To solve the limit \( \lim _{x \rightarrow 0} \frac{\tan 5x}{\sin 9x} \), we can use the known limit \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \) and the fact that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Rewriting the limit, we have: \[ \lim _{x \rightarrow 0} \frac{\tan 5x}{\sin 9x} = \lim _{x \rightarrow 0} \frac{\sin 5x}{\cos 5x \cdot \sin 9x} \] Next step is to express \( \sin 9x \) in terms of its limit: \[ \lim _{x \rightarrow 0} \frac{\sin 5x}{\sin 9x} = \lim _{x \rightarrow 0} \frac{\frac{\sin 5x}{5x}}{\frac{\sin 9x}{9x}} \cdot \frac{5x}{9x} \] Using the known limit, we have \( \lim_{x \rightarrow 0} \frac{\sin 5x}{5x} = 1 \) and \( \lim_{x \rightarrow 0} \frac{\sin 9x}{9x} = 1 \). Therefore, we get: \[ \lim _{x \rightarrow 0} \frac{\sin 5x}{\sin 9x} = \lim _{x \rightarrow 0} \frac{5x/9x}{1} = \frac{5}{9} \] Now, substituting back into the limit, we have: \[ \lim _{x \rightarrow 0} \frac{\tan 5x}{\sin 9x} = \lim _{x \rightarrow 0} \frac{\sin 5x}{\sin 9x \cos 5x} = \frac{5}{9} \cdot \frac{1}{1} = \frac{5}{9} \] Thus, the limit is: \[ \lim _{x \rightarrow 0} \frac{\tan 5x}{\sin 9x} = \frac{5}{9} \] So, the correct choice is: A. \( \lim _{x \rightarrow 0} \frac{\tan 5 x}{\sin 9 x} = \frac{5}{9} \)