From Saff and Snider's Fundamentals of Complex Analysis, section 4.3, problem 10.
Verify that if $f$ is continuous at $z$, then $\lim_{\Delta z \to 0} \int_{0}^{1}f(z+t\Delta z) dt = f(z)$.
Attempted solution:
$f(z)$ can be written as $\int_{0}^{1}f(z) dt$ so
$$\lim_{\Delta z \to 0} \int_{0}^{1}f(z+t\Delta z) dt = f(z) \iff $$
$$\lim_{\Delta z \to 0} \bigg[ \int_{0}^{1}f(z+t\Delta z) dt - f(z) \bigg] = \lim_{\Delta z \to 0} \int_{0}^{1} [f(z+t\Delta z) - f(z) ]dt = 0 .$$
Let $$M = \max_{t \in [0,1]} \lvert f(z+t\Delta z) - f(z) \rvert .$$
$$0 \leq \bigg| \int_{0}^{1}f(z+t\Delta z) - f(z) dt \bigg| \leq \int_{0}^{1} \lvert f(z+t\Delta z) - f(z) \rvert dt \leq M .$$
Now I think I need to show that taking $\Delta z \to 0$ implies $M \to 0$ but I don't know how. I think I have to employ the continuity of $f$, since I haven't used that yet.
If anyone could help me finish this proof, I would very much appreciate it.
You almost have it. Fix $\varepsilon>0$. Then there exists $\delta>0$ such that if $|h|<\delta$, $|f(z+h) - f(z) |<\varepsilon$. In particular, if $|\Delta z|<\delta$, you may replace $M$ above with $\varepsilon$, and so you're done.