Find the value of the following limit.
![\[ \lim_{x \to 0} \frac{ x \cot x - 1}{x^2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7d47fe99f6af673b61ac4847be74afe1_l3.png "Rendered by QuickLaTeX.com")
First, we write 
and then multiply inside the derivative by 
(which is alright since 
). Therefore we have,
![\begin{align*} \lim_{x \to 0} \frac{ x \cot x - 1}{x^2} &= \lim_{x \to 0} \left( \frac{ \frac{x \cos x}{\sin x} - 1}{x^2} \cdot \frac{\sin x}{x} \right) \\[9pt] &= \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \\[9pt] &= \lim_{x \to 0} \frac{ -x \sin x}{3x^2} &(\text{L'Hopital})\\[9pt] &= \lim_{x \to 0} \frac{- \cos x}{3} &(\text{L'Hopital}) \\[9pt] &= -\frac{1}{3}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cebf333600a843f5d794b6130ea7d024_l3.png "Rendered by QuickLaTeX.com")
Since this limit exists, the application of L’Hopital’s rule was justified.
Find the value of the following limit.
First, we write and then multiply inside the derivative by (which is alright since ). Therefore we have,
Since this limit exists, the application of L’Hopital’s rule was justified.