How could I evaluate this limit without using L’Hospital Rule

Question

$$\lim_{x\to 0}\int_0^x\frac{\sin t}{xt}\,dt$$

Answer 1

$$\int_0^x\frac{\sin(t)}{xt}\,dt=x\cdot\left(x^{-1}\frac{\sin(\xi)}{\xi}\right),\,\xi\in[0,x]=\frac{\sin(\xi)}{\xi}=\operatorname{sinc}(\xi)$$

By the mean value theorem.

By the squeeze theorem:

$$\lim_{x\to0}\operatorname{sinc}(\xi),0\le\xi\le x=\lim_{x\to0}\operatorname{sinc}(x)=1$$

Since the point $\xi$ is squeezed to zero by the limit on $x$. Sinc is a name for the function $\frac{\sin(x)}{x}$.