$\lim_{n \rightarrow \infty} f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$ 
Let 
$$f_n(x) = n^2 \left( 1- \cos \frac{x^3 - 1}{n} \right)$$
Let M be the set of x s.t. $\lim_{n \rightarrow \infty} f_n(x)$
  exists. For each $x \in M$ let $f(x) = \lim_{n \rightarrow \infty} f_n(x)$. Then
  
  
*
  
*M is bounded above
  
*f(x) is even function
  
*f(x) has slant asymptote
  
*$f'(1) = 6$
  
*1-4 are false.
  

The only thing I see now is that $1 \in M$. Please, give me hints how to deal with this problem.

Update: I try to use L'Hopital's rule
$$\lim \left(x^{3}-1\right)\sin\frac{x^{3}-1}{n}/\frac{2}{n^{2}}\rightarrow \lim \left(x^{3}-1\right)^{2}\cos\frac{x^{3}-1}{n}/\frac{4}{n^{2}}=\infty$$
 A: Assuming that $x\neq 1$ (the limit is easy to calculate for $x=1$, you can calculate the limit by writing $f(x)$ as $$f(x) = \frac{1-\cos\frac{x^3-1}{n}}{\frac{1}{n^2}},$$
then applying L'Hospital's rule twice.
Alternatively, you can use Taylor's expansion of $\cos$ to cancel some terms out.
A: it is easy to see that 
$\cos \frac{x^{3}-1}{n} = 1 - \frac{(\frac{x^{3}-1}{n})^{2}}{2} + ... = 1 - \frac{x^{6} - 2x^{3} + 1}{2n^{2}} + ...$ so that $f(x) = n^{2} (1-\cos \frac{x^{3}-1}{n}) \rightarrow \frac{x^{6}}{2} - x^{3} + \frac{1}{2}$
so 1 - false, 2 - false, 3 - false, 4 - false
A: Hint: $1 - \cos(x) \approx\frac{x^2}{2}$ for small $x$. Thus,
$$ f_n(x) = \frac{1 - \cos(\frac{x^3 - 1}{n})}{\frac{1}{n^2}} \approx \frac{(x^3 - 1)^2}{2} $$
for large $n$. To make this argument rigorous, use Taylor's theorem or L'Hopital rule.
A: Make the change of variable $\frac{x^3-1}n=\epsilon$, so that
$$\lim_{n\to\infty}n^2\big(1-\cos\frac{x^3-1}n\big)=(x^3-1)^2\lim_{\epsilon\to0}\frac{1-cos\epsilon}{\epsilon^2}.$$
The latter limit is a constant that can be computed by L'Hospital ($\to\frac{\sin\epsilon}{2\epsilon}\to\frac{\cos\epsilon}2\to\frac12$).
[For rigor, limits to $0^+$ or $0^-$ should be used depending on the sign of $x^3-1$, but given that the function is even with respect to $\epsilon$, this makes no difference.]
