update typo

pages/Math416/Math416_L12.md

@@ -61,7 +61,7 @@ $$
 \end{aligned}
 $$
 
-Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{z\toz_1}f(z)=f(z_1)$.
+Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{z\to z_1}f(z)=f(z_1)$.
 
 There exists a $\delta>0$ such that $|z-z_1|<\delta$ implies $|f(z)-f(z_1)|<\epsilon$.
 
@@ -71,7 +71,7 @@ $$
 |I|\leq\frac{1}{z_1-z_0}\int_{z}^{z_1}|f(\xi)-f(z_1)|d\xi<\frac{\epsilon}{z_1-z_0}\int_{z}^{z_1}d\xi=\epsilon
 $$
 
-So $I\to 0$ as $z_1\toz$.
+So $I\to 0$ as $z_1\to z$.
 
 Therefore, $g'(z_1)=f(z_1)$ for all $z_1\in U$.