Find $\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$ Find $$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx$$
One easy step is replacing $\sin t = t$, since $\lim_{t \rightarrow 0} \sin t / t = 1$. How do I continue with the rest?
 A: Fix $t\neq 0$. By the Mean Value Theorem for Integrals, there is some $c_t$ between  $0$ and $t$ such that
$$
\int_0^t\sqrt{1+\sin\left(x^2\right)}dx=t\sqrt{1+\sin\left(c_t^2\right)}
$$
Therefore,
$$\int_0^t\frac{\sqrt{1+\sin\left(x^2\right)}}{\sin(t)}dx=\frac{t}{\sin(t)}\sqrt{1+\sin\left(c_t^2\right)}$$
If $t\to 0$, the quotient tends to $1$, while the root tends to $1$, so the limit is $1$.
For a (marginally different) approach, let $f(t):=\int_0^t\sqrt{1+\sin\left(x^2\right)}dx$ and apply L'Hospital's Rule and the Fundamental Theorem of Calculus to
$$
\lim_{t\to 0} \frac{f(t)}{\sin(t)}
$$
A: Extract the first order terms from $\int_0^t \sqrt{1+\sin(x^2)} dx $.
Expanding binomially, $$\sqrt{1+\sin(x^2)} \approx 1+\frac 12 \sin (x^2)\\ \approx 1+\frac 12 x^2 \\ \approx 1$$ That means $$\int_0^t \sqrt{1+\sin(x^2)} dx \approx \int _0^t dx =t$$ and $$\lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx = \lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{t} dx  =1$$
A: Use the fundamental theorem of calculus and L'Hospital rule
$$\lim_{t \rightarrow 0} \int_{0}^{t} \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx=\lim_{t \rightarrow 0} \frac{\frac d{dt}\,\int_{0}^{t}\sqrt{1+\sin(x^2)}\,dx}{\frac d{dt}\,\sin(t) }$$ that is to say
$$\lim_{t \rightarrow 0} \frac{\sqrt{1+\sin(t^2)}}{\cos(t)}$$
A: The mean of a function over a domain $x\in[a,b]$ is given by:
$$\bar{f}=\frac{1}{b-a}\int_a^bf(x)dx$$
so if we have a domain of $x\in[0,t]$ then we can say:
$$\bar{f}=\frac1t\int_0^tf(x)dx$$
and as $t\to0$ the interval we are finding the average over is getting smaller until it is down to the point of $x=0$ and so we can say:
$$\lim_{t\to0}\int_0^t\sqrt{1+\sin(x^2)}dx=\lim_{t\to0}t\sqrt{1+\sin(t^2)}$$
and so for your integral you have:
$$\lim_{t\to0}\frac{t\sqrt{1+\sin(t^2)}}{\sin(t)}$$
and we can break this up and it is known that:
$$\lim_{t\to0}\frac{\sin(t)}{t}=1$$
so we have:
$$\lim_{t\to0}\sqrt{1+\sin(t^2)}=1$$
A: For small $t$, we know that $\sin (t)=t+\mathcal{O}(t^3)$, meaning that $\frac{1}{\sin(t)}=\frac{1}{t}\left[1+\mathcal{O}(t^2)\right]$ . Your limit is
$$\lim_{t\to\ 0}\frac{\int_0^t \sqrt{1+\sin(x^2)}\,dx - \int_0^0 \sqrt{1+\sin(x^2)}\,dx}{t}\left[1+\mathcal{O}(t^2)\right].$$
Hence, the limit is also
$$\frac{d}{dt}\Bigg|_{t=0}\int_0^t\sqrt{1+\sin(x^2)}\,dx=1.$$
