Multiple and iterated limits of functions of several variables.

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

The number $A$ is called the limit of the function $u=f(P)$ as the point $P(x_1, x_2, ..., x_n)$ tends to the point $P_0(a_1, a_2, ..., a_n)$, if for any $\varepsilon>0$ there exists $\delta>0$ such that from the condition $$0<\rho(P, P_0)=\sqrt{(x_1-a_1)^2+...+(x_n-a_n)^2}<\delta$$ it follows $$|f(x_1, x_2,..., x_n)-A|<\varepsilon.$$ In this case, it is written as: $$A=\lim\limits_{P\rightarrow P_0}f(P)=\lim\limits_{x_1\rightarrow a_1,...,x_n\rightarrow a_n}f(x_1, x_2, ..., x_n).$$

Examples:

Find the limits:

7.32. $\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy}{3-\sqrt{xy+9}}.$

Solution:

$$\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy}{3-\sqrt{xy+9}}=\left[\frac{0}{0}\right]=\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy(3+\sqrt{xy+9})}{(3-\sqrt{xy+9})(3+\sqrt{xy+9})}=$$ $$=\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy(3+\sqrt{xy+9})}{(9-{xy+9})}=\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy(3+\sqrt{xy+9})}{-xy}=$$ $$=-\lim\limits_{x\rightarrow 0, y\rightarrow 0}(3+\sqrt{xy+9})=-6.$$

Answer: $-12.$

7.34. $\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{\sin xy}{y}.$

Solution.

$$\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{\sin xy}{y}=\left[\frac{0}{0}\right]=[\sin xy\sim xy, \quad(x\rightarrow 0, y\rightarrow 0)]=\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{xy}{y}=\lim\limits_{x\rightarrow 0, y\rightarrow 0}x=0.$$

Answer: $0.$

7.36. $\lim\limits_{x\rightarrow \infty, y\rightarrow \infty}(x^2+y^2)\sin\frac{1}{x^2+y^2}.$

Solution.

$$\lim\limits_{x\rightarrow \infty, y\rightarrow \infty}(x^2+y^2)\sin\frac{1}{x^2+y^2}=\left[\infty\cdot 0\right]=$$ $$=\left[\sin\frac{1}{x^2+y^2}\sim \frac{1}{x^2+y^2}, \quad(x\rightarrow \infty, y\rightarrow \infty)\right]=\lim\limits_{x\rightarrow \infty, y\rightarrow \infty}\frac{x^2+y^2}{x^2+y^2}=1.$$

Answer: $1.$

7.37. Show that as $x\rightarrow 0$ and $y\rightarrow 0$, the function $z=\frac{x}{y-x}$ can approach any limit, provide examples of approximating the point $(x, y)$ to the point $(0, 0)$ such that $\lim z=3$, $\lim z=2$, and $\lim z=-2$.

Solution.

For any limit $A$, we can choose a subsequence of points ${M_k(\frac{A}{k},\frac{A+1}{k})}$ converging to the point $(0, 0)$ as $k\rightarrow \infty$. Then, $$z(M_k)=\frac{\frac{A}{k}}{\frac{A+1}{k}-\frac{A}{k}}=A.$$

Let's consider the subsequence of points ${M_k(\frac{3}{k},\frac{4}{k})}$ converging to the point $(0, 0)$ as $k\rightarrow \infty$. Then, $$\lim\limits_{x\rightarrow0, y\rightarrow 0}z=\lim\limits_{k\rightarrow\infty}z(M_k)=\lim\limits_{k\rightarrow\infty}\frac{\frac{3}{k}}{\frac{4}{k}-\frac{3}{k}}=3.$$

For the subsequence of points ${M_k(\frac{2}{k},\frac{3}{k})}$ converging to the point $(0, 0)$ as $k\rightarrow \infty$, we have: $$\lim\limits_{x\rightarrow0, y\rightarrow 0}z=\lim\limits_{k\rightarrow\infty}z(M_k)=\lim\limits_{k\rightarrow\infty}\frac{\frac{2}{k}}{\frac{3}{k}-\frac{2}{k}}=2.$$

For the subsequence of points ${M_k(\frac{-2}{k},\frac{-1}{k})}$ converging to the point $(0, 0)$ as $k\rightarrow \infty$, we have: $$\lim\limits_{x\rightarrow0, y\rightarrow 0}z=\lim\limits_{k\rightarrow\infty} z(M_k)=\lim\limits_{k\rightarrow\infty}\frac{\frac{-2}{k}}{\frac{-1}{k}-\frac{-2}{k}}=-2.$$

Answer: The function can approach any limit.

7.38. To show that for the function $f(x, y)=\frac{x-y}{x+y}$ there is no limit as $\lim\limits_{x\rightarrow 0, y\rightarrow 0},$ compute the repeated limits. $$\lim\limits_{x\rightarrow 0}\left(\lim\limits_{y\rightarrow 0} f(x, y)\right),\,\,\lim\limits_{y\rightarrow 0}\left(\lim\limits_{x\rightarrow 0} f(x, y)\right).$$

Solution.

$$\lim\limits_{x\rightarrow 0}\left(\lim\limits_{y\rightarrow 0} f(x, y)\right)=\lim\limits_{x\rightarrow 0}\left(\lim\limits_{y\rightarrow 0}\frac{x-y}{x+y}\right)=\lim\limits_{x\rightarrow 0}\frac{x-0}{x+0}=1.$$

$$\lim\limits_{y\rightarrow 0}\left(\lim\limits_{x\rightarrow 0} f(x, y)\right)=\lim\limits_{y\rightarrow 0}\left(\lim\limits_{x\rightarrow 0}\frac{x-y}{x+y}\right)=\lim\limits_{x\rightarrow 0}\frac{0-y}{0+y}=-1.$$

Since the repeated limits are different, the limit of the function $\lim\limits_{x\rightarrow 0, y\rightarrow 0}f(x, y)$ does not exist.

Answer: The limit does not exist.

7.40. Determine whether the function $\sin\ln(x^4+y^2)$ has a limit as $x\rightarrow 0, y\rightarrow 0?$

Solution.

Consider the partial sequence of points $\{M_k(\frac{1}{\sqrt[4]{2k}},\frac{1}{\sqrt{2k}})\}$ converging to the point $(0, 0)$ as $k\rightarrow \infty.$ Then $$\lim\limits_{x\rightarrow 0, y\rightarrow 0}z=\lim\limits_{k\rightarrow\infty}z(M_k)=\lim\limits_{k\rightarrow\infty}\sin\ln\left(\frac{1}{2k}+\frac{1}{2k}\right)=\lim\limits_{k\rightarrow\infty}\sin\ln\frac{1}{k}.$$

Since $\lim\limits_{k\rightarrow\infty}\ln\frac{1}{k}=-\infty,$ then $\lim\limits_{k\rightarrow\infty}\sin\ln\frac{1}{k}$ does not exist.

Answer: The function does not have a limit.

Homework:

7.33. $\lim\limits_{x\rightarrow 0, y\rightarrow 0}\frac{\sin xy}{xy}.$

7.35. $\lim\limits_{x\rightarrow 0, y\rightarrow 0}(1+x^2+y^2)^{\frac{1}{x^2+y^2}}.$

7.39. Show that for the function $f(x, y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ the repeated limits exist and are equal to each other. $$\lim\limits_{x\rightarrow 0}\left(\lim\limits_{y\rightarrow 0} f(x, y)\right),\,\,\lim\limits_{y\rightarrow 0}\left(\lim\limits_{x\rightarrow 0} f(x, y)\right),$$ тем не менее $\lim\limits_{x\rightarrow 0, y\rightarrow 0}$ does not exist.

7.41. Determine whether the function $\frac{x^2+y^4}{x^4+y^2}$ has a limit as $x\rightarrow \infty, y\rightarrow \infty.$

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