Math Help - limit help, please check my working

1. limit help, please check my working

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

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

is my method correct,

differentiate top and bottom,

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

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

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

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

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

IS this correct?

Thank you.

2. Re: limit help, please check my working

Yes! That is correct.

thank you.