I'm going to show the answer to this question in two different ways. - The first is perhaps more obvious but the second is much more elegant.

Taylor series expansion:

Using Taylor expansion (or your trusty A level formula sheet) you can show that sin(x) = x - x^3/3! + x^5/5! + Re ( (-i)^n * x^n / n! )

Thus dividing through by x:

sin(x)/x = 1 - x^2/3! + x^4/5! +...

if we then replace x by 0:

sin(0)/0 = 1 - 0 + 0 +... where ... here is all 0.

thus sin(0) / 0 = 1.

The other much faster way of doing this is using l'Hopital's rule which states that for a limit lim (f(x)/g(x)) = lim (f'(x) / g'(x)) for the same limit.

Thus lim[x-> 0] (sin(x) / x) = lim[x->0] (cos x / 1) = 1.