Complex limit question 
  
*
  
*Calculate $\displaystyle\lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}}$.  
  
*Calculate $\displaystyle\lim_{x\to 0} \frac{x^2e^{x^4}-\sin{(x^2)}}{1-\cos{(x^3)}}$.  
  

For the first question I got \begin{equation*}
\begin{split}
\lim_{x\to\infty} \frac{x^2e^{-4x}+xe^{-3x}}{e^{-2x}+xe^{-3x}} &= \lim_{x\to\infty} \frac{xe^{-3x}\left(xe^{-x}+1\right)}{e^{-2x}\left(xe^{-x}+1\right)} \\
&= \lim_{x\to\infty} \frac{xe^{-3x}}{e^{-2x}} \\
&= \lim_{x\to\infty} \frac{x}{e^x}.
\end{split}
\end{equation*}
This has an indeterminate form $\frac{\infty}{\infty}$. So we can use L'Hopital's rule to get $\lim_{x\to\infty} \frac{x}{e^x} = \lim_{x\to\infty} \frac{1}{e^x} = 0$. 
For the second question I tried L'Hopital's rule but that didn't work and I also tried multiplying by the conjugate $\left(1+\cos{(x^3)}\right)$. Any help would be great!!!

Near $0$, you have\begin{align}x^2e^{x^4}-\sin(x^2)&=x^2\left(1+x^4+\frac{x^8}{2!}+\cdots\right)-\left(x^2-\frac{x^6}6+\cdots\right)\\&=\frac76x^6+O(x^7)\end{align}and$$1-\cos(x^3)=\frac{x^6}2+O(x^7)$$and therefore$$\lim_{x\to0}\frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}=\frac{7/6}{1/2}=\frac73.$$
