Find the value of the following limit.
![\[ \lim_{x \to 0} \frac{(2-x)e^x - x - 2}{x^3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f7a7d0c21cea8ed36e2b28c6fbe93726_l3.png "Rendered by QuickLaTeX.com")
We apply L’Hopital’s rule three times,
![\begin{align*} \lim_{x \to 0} \frac{(2-x)e^x - x - 2}{x^3} &= \lim_{x \to 0} \frac{(2-x)e^x - e^x - 1}{3x^2} \\[9pt] &= \lim_{x \to 0} \frac{(2-x)e^x - 2e^x}{6x} \\[9pt] &= \lim_{x \to 0} \frac{(2-x)e^3 - 3e^x}{6} \\[9pt] &= -\frac{1}{6}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1fafe57037906106d501426dc938bd62_l3.png "Rendered by QuickLaTeX.com")
Since this final limit exists, the chain of applications of L’Hopital’s was justified.
Find the value of the following limit.
We apply L’Hopital’s rule three times,
Since this final limit exists, the chain of applications of L’Hopital’s was justified.