b) Evaluate: \( \lim _{x \rightarrow \theta} \frac{x \tan \theta-\theta \tan x}{-\theta} \)

To evaluate the limit \[ \lim _{x \rightarrow \theta} \frac{x \tan \theta - \theta \tan x}{-\theta}, \] let's first rewrite the limit in a more manageable form. We can factor out \(-\frac{1}{\theta}\): \[ \lim _{x \rightarrow \theta} \frac{x \tan \theta - \theta \tan x}{-\theta} = -\frac{1}{\theta} \lim _{x \rightarrow \theta} (x \tan \theta - \theta \tan x). \] As \(x\) approaches \(\theta\), both \(x \tan \theta\) and \(\theta \tan x\) approach \(\theta \tan \theta\). Thus, we can observe that the limit has an indeterminate form (i.e., \(0\) over \(0\)), so we can apply L'Hôpital's Rule. Now we need to differentiate the numerator and denominator: 1. The derivative of the numerator \(x \tan \theta - \theta \tan x\) with respect to \(x\) is: \[ \tan \theta - \theta \sec^2 x. \] 2. The derivative of \(-\theta\) with respect to \(x\) is \(0\). Applying L'Hôpital's Rule, we have: \[ \lim_{x \to \theta} \frac{x \tan \theta - \theta \tan x}{-\theta} = -\frac{1}{\theta} \lim_{x \to \theta} \left( \tan \theta - \theta \sec^2 x \right). \] Now we calculate this limit: \[ \lim_{x \to \theta} \left( \tan \theta - \theta \sec^2 x \right) = \tan \theta - \theta \sec^2 \theta. \] Putting this into our earlier limit gives: \[ -\frac{1}{\theta} \left( \tan \theta - \theta \sec^2 \theta \right). \] Thus, our limit becomes: \[ \lim _{x \rightarrow \theta} \frac{x \tan \theta - \theta \tan x}{-\theta} = -\frac{1}{\theta} (\tan \theta - \theta \sec^2 \theta). \] And this is our final simplification. Hence, the limit evaluates to: \[ \lim _{x \rightarrow \theta} \frac{x \tan \theta - \theta \tan x}{-\theta} = -\frac{\tan \theta - \theta \sec^2 \theta}{\theta}. \]