find $\lim _{t \to \infty}t^3 \left(1-\{f'(t)\}^2 \right)=?$ I would appreciate if somebody could help me with the following problem
Q: $\forall t \in R$, $f(t)$: differential function in $ R$ and  satisfy 
$$\int_{t}^{f(t)}\sqrt{1+ 9x^4} dx =1 (0<t<f(t))$$
find $$\lim _{t \to \infty}t^3 \left(1-\{f'(t)\}^2 \right)=?$$
I find
$$t^3 \left(1-\{f'(t)\}^2 \right)=\frac{9\{f(t)\}^4 t^3- 9t^7}{1+9\{f(t)\}^4}$$
and ...
 A: I remember this problem, because it was a content of the entrance exam of Seoul National University (2008).
Let
$$
g(t)=\int_t^{f(t)}\sqrt{1+9x^4}dx,
$$
then
$$
g'(t)=f'(t)\sqrt{1+9f(t)^4}-\sqrt{1+9t^4}.
$$
Since $g(t)$ is constant, we get
$$
f'(t)=\frac{\sqrt{1+9t^4}}{\sqrt{1+9f(t)^4}}
$$
and so
$$
t^3(1-f'(t)^2)=\frac{9t^3(f(t)^4-t^4)}{1+9f(t)^4}
$$
By mean value theorem, there exists $c$ ($t<c<f(t)$) such that
$$
\frac{1}{f(t)-t}\int_t^{f(t)}\sqrt{1+9x^4}dx=\sqrt{1+9c^4}
$$
and so
$$
f(t)-t=\frac{1}{\sqrt{1+9c^4}}.
$$
Show that $\lim_{t\to\infty}\frac{f(t)}{t}=1$. Since $3x^2 \le \sqrt{1+9x^4}$,
$$
f(t)^3-t^3=\int_t^{f(t)}3x^2 dx \le \int_t^{f(t)} \sqrt{1+9x^4}dx=1.
$$
Thus
$$
0\le \frac{f(t)}{t}-1\le \frac{1}{tf(t)^2+t^2f(t)+t^3}\le \frac{1}{3t^3}
$$
By sandwich theorem, we get $\lim_{t\to\infty}\frac{f(t)}{t}=1$.
Therefore,
\begin{align}
\lim_{t\to\infty}\frac{9t^3(f(t)^4-t^4)}{1+9f(t)^4}&=\lim_{t\to\infty}\frac{9t^3(f(t)+t)(f(t)^2+t^2)}{(1+9f(t)^4)\sqrt{1+9c^4}}\\
&=\lim_{t\to\infty}\frac{9\left(\frac{f(t)}{t}+1\right)\left(\frac{f(t)^2}{t^2}+1\right)}{\left(\frac{1}{t^4}+\frac{9f(t)^4}{t^4}\right)\sqrt{\frac{1}{t^4}+\frac{9c^4}{t^4}}}\\
&=\frac{9\cdot 2\cdot 2}{9\cdot 3}\\
&=\frac{4}{3}.
\end{align}
