Chapter 15 - Differentiation in Several Variables - 15.2 Limits and Continuity in Several Variables - Exercises - Page 772: 29

$$0$$

Given $$\lim _{(x, y) \rightarrow(4,2)} \frac{y-2}{\sqrt{x^{2}-4}} $$ Since $ \dfrac{y-2}{\sqrt{x^{2}-4}} $ is continuous on $ ( 4,2)$, then by substitution, we get \begin{align*} \lim _{(x, y) \rightarrow(4,2)} \frac{y-2}{\sqrt{x^{2}-4}} &=\lim _{(x, y) \rightarrow(4,2)} \frac{2-2}{\sqrt{16-4}} \\ &=0 \end{align*}