Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize. Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,
First Solution :
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$
$=\lim\limits_{x\to 1}\dfrac{x+1}{1}=\dfrac{1+1}{1}=2$
And Second Solution with l'hopital: 
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}\overbrace{\longrightarrow}^{l'hopital}\lim\limits_{x\to 1}\dfrac{2x}{1}=2$
But...
Let's we take this function ,$f(x)=\dfrac{1-cos(x^6)}{x^{12}}$
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0/0$,
For this funciton couldn't simplification,I have just , graph of function and l'hôpital,
1-L'hôpital;
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=\lim\limits_{x\to 0}\dfrac{6.x^5.sin(x^6)}{12.x^{11}}=\underbrace{\lim\limits_{x\to 0}\dfrac{sin(x^6)}{x^6}}_1.\dfrac{1}{2}=\dfrac{1}{2}$
2-Graph of function;
Graph telling me ,$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0$
l'hôpital $\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=1/2$
What we do now?

And link of graph , if you can't see clearly,https://www.desmos.com/calculator/jfz4kslm4w

Hint:
$$
\begin{align}
\frac{1-\cos\left(x^6\right)}{x^{12}}
&=\frac{1-\cos\left(x^6\right)}{\sin^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac{1-\cos\left(x^6\right)}{1-\cos^2\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
&=\frac1{1+\cos\left(x^6\right)}\left(\frac{\sin\left(x^6\right)}{x^6}\right)^2\\
\end{align}
$$
