Show that $\lim_{(x,y)\to(1,2)} (x^2 + 2y) = 5$ using definition of limit.
Question
We must show that for any $\epsilon > 0$, there exists $\delta > 0$ such that
$$0 < \sqrt{(x-1)^2 + (y-2)^2} < \delta \implies |x^2 + 2y - 5| < \epsilon$$
Then,| $|f(x,y) - L|$ | $= |x^2 + 2y - 5|$ |
| $= |x^2 - 1 + 2(y - 2)|$ | |
| $= |(x-1)(x+1) + 2(y - 2)|$ |
Use the triangle inequality
$$|x^2 + 2y - 5| \le |x-1||x+1| + |2(y - 2)|$$
We assume that $\sqrt{(x-1)^2 + (y-2)^2} < \delta$ then by the definition of Euclidean distance:
$$|x-1| \le \sqrt{(x-1)^2 + (y-2)^2} < \delta$$ $$|y-2| \le \sqrt{(x-1)^2 + (y-2)^2} < \delta$$
Let's restrict $\delta \le 1$, then $|x-1| < 1 \implies x \in (0,2),$
then $|x+1| < 3$
So, $|x^2 + 2y - 5| \le 3|x-1| + 2|y - 2|$
Thus $|x^2 + 2y - 5| < 3\delta + 2\delta = 5\delta$
For any $\epsilon > 0$, choose $\delta = \min \left(1, \frac{\epsilon}{5}\right)$
Then if $\sqrt{(x-1)^2 + (y-2)^2} < \delta$ we have $|x^2 + 2y - 5| < 5\delta = 5 \cdot \frac{\epsilon}{5} = \epsilon$
So, $d((x,y), (1,2)) < \delta$ then $|x^2 + 2y - 5| < \epsilon$
Hence; $$\boxed{\lim_{(x,y)\to(1,2)} (x^2 + 2y) = 5}$$
