Example 1. Calculate the limits:

a) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\tan\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$, b) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, c) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x y}{\sqrt{4-x y}-2}$.
Solution. Note that the functions $\frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$ and $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ are undefined only at the point $(0,0)$, while $\frac{x y}{\sqrt{4-x y}-2}$ is undefined on the coordinate axes $x=0$ and $y=0$.

a) Transition to polar coordinates $x=r \cos \varphi, y=r \sin \varphi$, $\varphi \in[0,2 \pi)$. Then $x^{2}+y^{2}=r^{2}$ and $(x, y) \rightarrow(0,0)$ implies $r \rightarrow 0$. Using the corollary of the first remarkable limit, we get

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}=\left(\frac{0}{0}\right)=\lim _{r \rightarrow 0} \frac{\operatorname{tg} r^{2}}{r^{2}}=1
$$

b) If $(x, y) \rightarrow(0,0)$ along different rays $y=k x, k \in \mathbb{R}$, then

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{x^{2}-k^{2} x^{2}}{x^{2}+k^{2} x^{2}}=\frac{1-k^{2}}{1+k^{2}}
$$

The right-hand side of this equation depends on $k$, and therefore the limit of the given function as $(x, y) \rightarrow(0,0)$ does not exist (the limit must be unique).

c) Multiply the numerator and denominator by the conjugate of the denominator. We get

$$
\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x y(\sqrt{4-x y}+2)}{-x y}=-\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(\sqrt{4-x y}+2)=-4
$$

Note. From the definition of continuity, it follows that the functions

$$
f(x, y)=\left\{\begin{array}{l}
\frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} \quad \text { for }(x, y) \neq(0, y) \text { and }(x, y) \neq(x, 0) \\
1 \quad \text { for }(x, y)=(0, y) \text { or }(x, y)=(x, 0)
\end{array}\right.
$$

and

$$
g(x, y)=\left\{\begin{array}{l}
\frac{x y}{\sqrt{4-x y}-2} \quad \text { for }(x, y) \neq(0,0) \\
-4 \text { for }(x, y)=(0,0)
\end{array}\right.
$$

are continuous on the entire plane $O x y$, while the function $\varphi(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ cannot be extended at the point $(0,0)$ in such a way that it is continuous at this point.

## Exercises

Calculate the limits.

1. $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\operatorname{tg} x y}{y}$. 2. $\lim _{\substack{x \rightarrow 3 \\ y \rightarrow 7}} \lg (x+y)$.
2. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\sin x y}{x}$.
3. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}$
4. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}}(x+y) \sin \frac{1}{x} \cdot \cos \frac{1}{y}$.
5. $\lim _{\substack{x \rightarrow 1 \\ y \rightarrow 2}} \frac{2(x-1)(y-2)}{(x-1)^{2}+(y-2)^{2}}$
6. $\lim _{\substack{x \rightarrow \infty \\ y \rightarrow \infty}} \frac{x^{2}+y^{2}}{x^{4}+y^{4}}$
7. $\lim _{\substack{x \rightarrow \infty \\ y \rightarrow \infty}} \frac{x+y}{x^{2}-x y+y^{2}}$.
8. $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2} y}{x^{2}+y^{2}}$
9. $\lim _{\substack{x \rightarrow 1 \\ y \rightarrow 0}} \frac{\ln \left(1+e^{y}\right)}{\sqrt{x^{2}+y^{2}}}$
10. $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow-2}} \frac{2 x+y}{(x-2)^{2}+(y+2)^{2}}$
11. $\lim _{\substack{x \rightarrow-1 \\ y \rightarrow 2}} \frac{x^{2}+2 x+y^{2}-4 y+5}{x+2 y}$.

## Answers

1. 2. 2. 3. 3. 0. 4. Does not exist. 5. 0. 6. Does not exist. 7. 0. 8.0. 9. 0. 10. $\ln 2.11 .+\infty$. 12.0 .

## § 3. Partial Derivatives and Differential of a Function of Two Variables

$1^{\circ}$. The partial increments of the function $z=f(x, y)$ with respect to the independent variables $x$ and $y$ are the differences

$$
\Delta_{x} z=f(x+\Delta x, y)-f(x, y), \quad \Delta_{y} z=f(x, y+\Delta y)-f(x, y)
$$

where $\Delta x$ and $\Delta y$ are the increments of the independent variables $x$ and $y$.

The total increment of the function $z=f(x, y)$ is the difference

$$
\Delta z=f(x+\Delta x, y+\Delta y)-f(x, y)
$$

In general, the total increment does not equal the sum of the partial increments:

$$
\Delta z \neq \Delta_{x} z+\Delta_{y} z
$$

$2^{\circ}$. The partial derivative of the function $z=f(x, y)$ with respect to the variable $x$ or $y$ is the limit of the ratio of the corresponding partial increment $\Delta_{x} z$ or $\Delta_{y} z$ to the increment of the given variable, provided that the increment of the variable tends to zero.

The notations for partial derivatives are

$$
\begin{aligned}
& z_{x}^{\prime}=\frac{\partial z}{\partial x}=\frac{\partial f(x, y)}{\partial x}=f_{x}^{\prime}(x, y)=\lim _{\Delta x \rightarrow 0} \frac{\Delta_{x} z}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x} \\
& z_{y}^{\prime}=\frac{\partial z}{\partial y}=\frac{\partial f(x, y)}{\partial y}=f_{y}^{\prime}(x, y)=\lim _{\Delta y \rightarrow 0} \frac{\Delta_{y} z}{\Delta y}=\lim _{\Delta y \rightarrow 0} \frac{f(x, y+\Delta y)-f(x, y)}{\Delta y}
\end{aligned}
$$

$3^{\circ}$. When finding a partial derivative (differentiation) with respect to a certain variable, one uses the formulas and rules of differentiation of a function of one variable, treating the other variable as fixed, constant.

$4^{\circ}$. The partial derivative $z_{x}^{\prime}$ $\left(z_{y}^{\prime}\right)$ at a given point $\left(x_{0}, y_{0}\right)$ is equal to the tangent of the angle of inclination of the tangent to the curve obtained by the intersection of the surface

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-097.jpg?height=382&width=587&top_left_y=1311&top_left_x=1891)

Figure 8.2 $z=f(x, y)$ with the plane $y=y_{0}$ $\left(x=x_{0}\right), z_{x}^{\prime}\left(x_{0}, y_{0}\right)=\operatorname{tg} \alpha\left(z_{y}^{\prime}\left(x_{0}, y_{0}\right)=\operatorname{tg} \beta\right)$ (Figure 8.2).

$5^{\circ}$. The partial differentials of the function $z=f(x, y)$ are the quantities

$$
d_{x} z=z_{x}^{\prime} d x, \quad d_{y} z=z_{y}^{\prime} d y, \quad d x=\Delta x, \quad d y=\Delta y
$$

$6^{\circ}$. The total differential of the function $z=f(x, y)$ is the expression

$$
d z=d_{x} z+d_{y} z=z_{x}^{\prime} d x+z_{y}^{\prime} d y=\frac{\partial z}{\partial x} d x+\frac{\partial z}{\partial y} d y
$$

The total differential $d z$ represents the principal linear (with respect to $\Delta x$ and $\Delta y$) part of the total increment of the function $z=f(x, y)$.

## Examples with Solutions