Evaluate $\lim_{x\to \theta} \dfrac {x \cot \theta - \theta \cot x}{x-\theta}$ Evaluate $\lim_{x\to \theta} \dfrac {x \cot \theta - \theta \cot x}{x-\theta}$
My Attempt:
$$=\lim_{x\to \theta} \dfrac {x \cot \theta - \theta \cot x}{x-\theta}$$
It takes $\dfrac {0}{0}$ form when $x=\theta $
$$=\lim_{x\to \theta} \dfrac {x\cot \theta -\theta \cot \theta + \theta \cot \theta - \theta \cot x}{x-\theta }$$.
How do I proceed further?
 A: If you rewrite:
$$\begin{align}x \cot \theta - \theta \cot x & = x \cot \theta \color{red}{- \theta \cot \theta+\theta \cot \theta}  - \theta \cot x \\[5pt] 
& = \cot \theta \left( x- \theta\right)- \theta \left( \cot x-\cot \theta\right)\end{align}$$
Then:
$$\begin{align}\lim_{x\to \theta} \frac {x \cot \theta - \theta \cot x}{x-\theta}
& =\lim_{x\to \theta} \frac {\cot \theta \left( x- \theta\right)- \theta \left( \cot x-\cot \theta\right)}{x-\theta} \\[8pt]
& =\cot \theta-\theta\;\color{blue}{\lim_{x\to \theta} \frac {\cot x-\cot \theta}{x-\theta}} \\[6pt]
\end{align}$$
Now notice that the limit in blue is, by definition, the derivative of $\cot x$ at $\theta$.

Addition after comments. If you can't use derivatives but you have trigonometric identities and the standard $\sin$ limit, you can proceed on the blue limit above with the formula:
$$\cot x-\cot \theta = \frac{-\sin(x-\theta)}{\sin x \sin \theta}$$
Then:
$$\color{blue}{\lim_{x\to \theta} \frac {\cot x-\cot \theta}{x-\theta}} = 
\lim_{x\to \theta} \frac{-\sin(x-\theta)}{\left( x-\theta \right)\sin x \sin \theta} = -\csc^2\theta \; \color{red}{\lim_{x\to \theta} \frac{\sin(x-\theta)}{x-\theta}}$$
Where the red limit is $1$ by the standard limit; substitute $h=x-\theta$ to get it in standard form.
A: Using L'Hopista's rule gives
$$
\lim_{x\to\theta}\frac{x\cot\theta-\theta\cot x}{x-\theta}
\lim_{x\to\theta}\frac{\cot\theta-\theta\frac{1}{\sin^2x}}{1}
=\cot\theta-\frac{\theta}{\sin^2\theta}
$$
