Limit of a function with a defined integral I have the next limit:
\begin{equation}
\lim_{x\to\ 0}\displaystyle\frac{\displaystyle\int_0^{x^2}{\frac{1-\cos{t^2}+at^4}{t}}dt}{(1-\cos{(\frac{x}{3})})^4}
\end{equation}
I've tried to solve it by doing L'Hôpital, and then I had the variable $x$ everywhere, so I applied some infinitessimals equivalences and then I realised that the limit is $0$, and the value of $a$ doesn't matter (because when I simplified everywhere, the $x$ in the numerator has degree 8, and in the denominator has degree 7, so it went to $0$).
What I have say above, it's correct or I must obtain a different solution? Thanks!
 A: You've committed an error somewhere.  Having not seen your work, I can't identify where, though. 
Are you familiar with power series?  They provide one method for going through this problem.  Let me show you; if you aren't familiar with them, at least take it as confirmation that there's an error somewhere. :-)
Note that
$$
-\cos(t^2)+at^4=at^4-\sum_{n=0}^{\infty}\frac{(-1)^nt^{4n}}{(2n)!},
$$
so that
$$
\frac{1-\cos(t^2)+at^4}{t}=\frac{1}{t}\left(at^4-\sum_{n=1}^{\infty}\frac{(-1)^nt^{4n}}{(2n)!}\right)=at^3-\sum_{n=1}^{\infty}\frac{(-1)^nt^{4n-1}}{(2n)!}.
$$
Integrating yields
$$
\int_0^{x^2}\frac{1-\cos(t^2)+at^4}{t}\,dt=\frac{a}{4}x^8-\sum_{n=1}^{\infty}\frac{(-1)^nx^{8n}}{4n\,(2n)!}=\frac{1+2a}{8}x^8+O\left(x^{16}\right).
$$
Now, let's look at the denominator. We have
$$
1-\cos\left(\frac{x}{3}\right)=1-\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{3^{2n}(2n)!}=\frac{x^2}{18}+O(x^4),
$$
and therefore
$$
\left(1-\cos\left(\frac{x}{3}\right)\right)^4=\frac{x^8}{18^4}\left(1+O(x^2)\right).
$$
So, all told, we have
$$
\lim_{x\to0}\frac{\displaystyle\int_0^{x^2}\frac{1-\cos(t^2)+at^4}{t}\,dt}{\left(1-\cos(\frac{x}{3})\right)^4}=18^4\cdot\frac{1+2a}{8}=13122(1+2a).
$$
A: Proceed with L'hopital rule...: Differentiate both numerator and denominator and simplify to get:
$2x\cdot \dfrac{1-cosx^4 + ax^8}{x^2}\cdot \dfrac{1}{4(1-cos(\frac{x}{3}))^3\cdot \frac{1}{3}\cdot sin(\frac{x}{3})} = L$. Using: $sinx \sim_0 x$, $1 - cosx = 2sin^2(\frac{x}{2})\sim_0 2\cdot \left(\frac{x}{2}\right)^2$, and the substitution of $ y = x^4$, we get:
$L \sim_0 \dfrac{9\cdot 6^6}{16}\cdot \left(\dfrac{1 - cosy}{y^2} + \dfrac{ay^2}{y^2}\right) \to \dfrac{9\cdot 6^6}{16}\cdot \left(\dfrac{1}{2} + a\right)$
