Question

How do you evaluate `[tan^2(3x)] / (2x)` as x approaches 0?

Answer 1

The requested limit can be written in the form

`lim_(x->0) (9x)/2 *[(sin 3x)/(3x)*1/(cos 3x)]^2`

Hence

\begin{equation}\lim_{{x}\to0}~{\frac{9{x}}{2}}~.{{\left[{\frac{{\operatorname{sin}~{3{x}}}}{3{x}}}.{\frac{1}{{\operatorname{cos}~{3{x}}}}} \right]}^{2}}=\lim_{{x}\to0}~{\frac{9{x}}{2}}~.{{\left[\lim_{{x}\to0}~~{\frac{{\operatorname{sin}~{3{x}}}}{3{x}}}.\lim_{{x}\to0}~{\frac{1}{{\operatorname{cos}~{3{x}}}}} \right]}^{2}}= \ =\lim_{{x}\to0}~{\frac{9{x}}{2}}~.{\left[1.{\frac{1}{1}} \right]}=0\end{equation}

Finally

`lim_(x->0) (9x)/2 *[(sin 3x)/(3x)*1/(cos 3x)]^2=0`

Note that `tan 3x=(sin 3x)/(cos 3x)` and `lim_(x->0) (sinx)/x =1`