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

1 Answer
Konstantinos Michailidis
Mar 25, 2016

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#

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