I have computed limits rigorously before but I have never come across an example where the variable is located in the upper limit of a definite integral. The exact question is attached.
It's shown below as part $\rm(iv)$. I've managed to show the parts before this. Also, would the second part be done by using Leibniz' rule?
$\rm(iv)$ You may assume that if $K>0$ is a constant and $$d=\dfrac{\log(S/K)+(r-\tfrac{1}{2}\sigma^2)(T-t)}{\sqrt{\sigma^2(T-t)}},\quad\Phi(x)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-p^2/2}\,dp,$$ then $$V(S,t)=e^{-r(T-t)}\Phi(d)$$ is a solution of the Black-Scholes equation in $(1)$, for $S>0$, $t<T$, $\sigma>0$. For this particular $V$, and assuming $S>0$ and $t<T$, find $$f(S)=\lim_{t\to T^{-}}V(S,t)\quad\text{and}\quad\rho(S,t)=\dfrac{\partial V}{\partial r}(S,t).$$
The integral in question is just finding the cumulative Normal probability. If $x \rightarrow\infty$, the integral goes to 1. The calculation of d just seems to be finding the z-value for a standard $N(0,1)$