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?
 A: It is best to first establish the well known limit $$\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{1}{3}\tag{1}$$ This is easily done via the application of L'Hospital's Rule or Taylor series expansions.
The exponent $12$ is used here to intimidate students and one just needs to replace it with a generic symbol $n$. Next we note that for any positive integer $n$ we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\frac{\tan x-x} {x^3}\cdot\sum_{i=1}^{n}\frac{\tan^{i-1}x}{x^{i-1}} =\frac{n}{3}$$ The limit in question is $f(12)=4$.
Thus if one wishes to use L'Hospital's Rule then just a single application of the rule is fine. But one must always use a certain amount of algebraic manipulation before applying the rule.

In case $n$ is not a positive integer then it is best to make use of another standard limit $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{2}$$ along with limit $(1)$. Thus we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\dfrac{\left(\dfrac{\tan x} {x} \right) ^n-1}{\dfrac{\tan x} {x} - 1}\cdot\dfrac{\dfrac{\tan x} {x} - 1}{x^2}\\=\lim_{t\to 1}\frac{t^n-1}{t-1}\cdot\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{n}{3}$$ Here we have used the substitution $t=(\tan x) /x$ so that $t\to 1$ as $x\to 0$.
A: Hint
The simplest would be to use Taylor series
$$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ 
$$ \frac{\tan^{12}(x)-x^{12}}{x^{14}}=\frac{x^{12}\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -x^{12}}{x^{14} }=\frac 1 {x^2}\left(\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -1\right)$$ Now, use the binomial expansion.
Even simpler, remember that, for small $\epsilon$, $(1+\epsilon)^n \sim 1+n\epsilon$
A: $\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}
$
