how to find the limit of: $\,\,a_{n+1}=\int\limits _{0}^{a_{n}}\sin\left(t^{2}\right)\,dt\quad,\quad a_{1}=1$? I would appreciate it if someone described a method for solving this. This is not a homework question, just something I stumbled upon and am curious about. Thanks.
 A: It seems fairly obvious that
$$\left|a_{n+1}\right| \le \int_0^{a_n} \left|\, \sin(t^2)\,\right|\,dt \le \int_0^{a_n}\sin(1)\,dt = \sin(1)\cdot a_n,$$
so by the Squeeze Theorem, the limit ought to be zero.
Details: 
(1) $\displaystyle\left|\int_0^{a_n} \sin(t^2)\,\,dt\right| \le \int_0^{a_n} \left|\, \sin(t^2)\,\right|\,dt$ is a basic property of integrals.
(2) $\sin(t^2)$ is increasing and positive on $[0,1]$, so $\sin(t^2)\le \sin(1^2)=\sin(1)$, for all $t\in [0,a_n] \subseteq [0,1]$.
(3) $\displaystyle\int_0^{a_n}\sin(1)\,dt = \sin(1)\cdot a_n = \sin(1)\cdot \left| a_n \right|$
Repeating this formula $n$ times yields
$$0 \le \left|a_{n+1}\right| \le (\sin (1))^n\cdot a_1.$$
The limit of the right-hand side is zero, because $\sin(1)<1$. The limit of the left-hand side is zero for obvious reasons. Therefore, by the discrete version of the Squeeze Theorem, $\displaystyle\lim_{n\to\infty} a_n = 0$. QED.
A: First you show $a_n$ is non-negative and bounded above by $1$ by induction on $n$. $a_1 = 1 \leq 1$, thus assume $a_n \leq 1$, then $a_{n+1} = \displaystyle \int_{0}^{a_n} \sin(t^2)dt\leq \displaystyle \int_{0}^{1} \sin(t^2)dt\leq \displaystyle \int_{0}^1 1dt = 1$. Next you prove $a_n \geq 0$ by induction on $n$ as well. $a_1 = 1 > 0$, and assume $a_n \geq 0$, then $a_{n+1} = \displaystyle \int_{0}^{a_n} \sin(t^2)dt \geq \displaystyle \int_{0}^{a_n} 0 dt = 0$ ( this is true because $0 \leq t \leq a_n \leq 1 \Rightarrow \sin(t^2) \geq 0$), and also $a_{n+1} \leq \displaystyle \int_{0}^{a_n} 1dt=a_n$, thus the sequence $a_n$ is decreasing and is bounded below by $0$, hence converges to $L\geq 0$. Passing to limit the above equation we have: $L = \displaystyle \int_{0}^L \sin(t^2)dt$. Now if $L > 0 \Rightarrow L < \displaystyle \int_{0}^L \sin(1)dt< L$, contradiction. Thus $L = 0$.
