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?

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 $$\lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{\sin t} dx = \lim_{t\to 0} \int_0^t \frac{\sqrt{1+\sin(x^2)}}{t} dx  =1$$
