From part (a) we have

$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{4kt} } f(y) dy $

Rewrite the formula as:

$u(x,t) = \frac{1}{\sqrt{2\pi } \sqrt{2kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{2 \sqrt{2kt}^2 } } f(y) dy $

We see this is a normal density of random variable x centered at y with standard error $ \sqrt{2kt}$ . When $t \to 0$, the standard error approach to 0 as well. The random variable x approaches to a deterministic form. So $lim_{t \to 0} u(x,t) = f(y) $.

I think the argument can be written more accurate if we take the functional form of f(x) in the whole real line. But then we are not directly prove the result based on the formula of u(x,t) we get from part (a).