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?
Best Answer
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.