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?

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
$$
\lim_{t\to 0} \frac{f(t)}{\sin(t)}
$$
