Examples

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In every case we assume that the conditions for the application of L'Hopital's rule have been satisfied.

The original quotient:

$\begin{displaymath} \frac{\tan(x) -x}{x^3}.\end{displaymath}$

The quotient of the derivatives:

$\begin{displaymath} \frac{\sec^2(x) -1}{3x^2} = \frac{1}{3\cos^2(x)}\left(\frac{\sin(x)}{x}\right)^2 \end{displaymath}$

has a limit of 1/3 has x tends to . Therefore

$\begin{displaymath} \lim_{x\rightarrow 0}\frac{\tan(x) -x}{x^3} = \frac{1}{3}\end{displaymath}$

The original quotient:

$\begin{displaymath} \frac{\arctan(x)}{\arcsin(x)}.\end{displaymath}$

The quotient of the derivatives:

$\begin{displaymath} \frac{\frac{1}{1+x^2}}{\frac{1}{\sqrt{1-x^2}}} = \frac{\sqrt{1-x^2}}{1+x^2}\end{displaymath}$

clearly has a limit of 1 as x tend to . Therefore

$\begin{displaymath} \lim_{x\rightarrow 0}\frac{\arctan(x)}{\arcsin(x)} = 1.\end{displaymath}$

David G Radcliffe
8/18/1998