First take a look at the following diagram, showing three areas arranged in order of magnitude.

The shaded area on the left (the smallest of the three) is a triangle with height of length $\sin h$ and base of length $\cos h.$ Therefore, its area is $\frac{(\cos h)(\sin h)}{2}.$
The pink shaded area (the next-smallest) is a circular segment compromising the fraction $\frac{h}{2π}$ of the entire disc. Since the area of a disc of radius $1$ is $π,$ the area in question is

$\frac{h}{2π}\ .\ π = \frac{h}{2}$

The shaded area on the right (the largest of the three) is a triangle with height of length $\tan h$ and base of length $1.$ Therefore, its area is $\frac{(1)(\tan h)}{2} = \frac{\tan h}{2}.$

Substituting these three areas therefore gives the inequality

Finally, let $h$ approach zero. As is does, the quantities on either end approach $1.$ Therefore, since the ratio $\frac{\sin h}{h}$ is sandwiched between two quantities approaching $1,$ it also approaches $1.$

We are now done with the first limit we promised to compute.

Proof of (2)

For the second limit, we use a trigonometric identity and a little algebra:

The first term in this product is the limit we computed above, and has the value of $1.$ The second term approaches $\frac{0}{(1+1)} = 0.$ Therefore, the product approaches $(1)(0) = 0,$ as required.