Evaluate $\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $ $$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $$
I tried writing it as  $\lim\limits_{x \to 0} \dfrac{\dfrac{\tan^{12}x}{x^{12}}-1}{x^{2}} $ thinking that if I applied L'hospital two times I would reach my result but with no luck. Any suggestions or tricks I could use?

$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}}
$
$\begin{array}\\
\lim_{x \to 0} \dfrac{\tan^{12}x-x^{12}}{x^{14}}
&=\lim_{x \to 0} \dfrac{(\tan(x)/x)^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\sin(x)/(x\cos(x))^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{\sin(x)}{x\cos(x)})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x(1-x^2/2+O(x^4))})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x-x^3/2+O(x^5))})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{1-x^2/6+O(x^4)}{1-x^2/2+O(x^4)})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(1+x^2/3+O(x^4))^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{1+4x^2+O(x^4)-1}{x^{2}}\\
&=\lim_{x \to 0} 4+O(x^2)\\
&=4\\
\end{array}
$
