Use L’Hopitals rule, with a substitution where necessary, to find

$\displaystyle \lim_{x \to 0} \frac{tanx}{tan3x} $

is my method correct,

differentiate top and bottom,

$\displaystyle \frac{sec^{2}x}{3sec^{2}3x} $

$\displaystyle sec^{2} x = tan^{2}x +1 $

$\displaystyle sec^{2}3x = tan^{2}3x +1 $

$\displaystyle \frac{tan^{2}x+1}{3tan^{2}3x +3} $

now evaluate the limit and it gives $\displaystyle \frac{1}{3} $ ?

IS this correct?

Thank you.