Multiple and iterated limits of functions of several variables.

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 $\epsilon >0$ 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 $$|f(x_1,x_2,...,x_n)-A|<\epsilon .$$ In this case, it is written as: $$A=\underset{P\to P_0}{lim}f(P)=\underset{x_1\to a_1,...,x_n\to a_n}{lim}f(x_1,x_2,...,x_n).$$

Examples:

Find the limits:

1. $\underset{x\to 0,y\to 0}{lim}\frac{xy}{3-\sqrt{xy+9}}.$

Solution:

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

Answer: $-12.$

2. $\underset{x\to 0,y\to 0}{lim}\frac{\sin xy}{y}.$

Solution.

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

Answer: $0.$

3. $\underset{x\to \mathrm{\infty },y\to \mathrm{\infty }}{lim}(x^2+y^2)\sin \frac{1}{x^2+y^2}.$

Solution.

$$\underset{x\to \mathrm{\infty },y\to \mathrm{\infty }}{lim}(x^2+y^2)\sin \frac{1}{x^2+y^2}=[\mathrm{\infty }\cdot 0]=$$ $$=[\sin \frac{1}{x^2+y^2}\sim \frac{1}{x^2+y^2},(x\to \mathrm{\infty },y\to \mathrm{\infty })]=\underset{x\to \mathrm{\infty },y\to \mathrm{\infty }}{lim}\frac{x^2+y^2}{x^2+y^2}=1.$$

Answer: $1.$

4. Show that as $x\to 0$ and $y\to 0$, 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 $limz=3$, $limz=2$, and $limz=-2$.

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 $k\to \mathrm{\infty }$. 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 $k\to \mathrm{\infty }$. Then, $$\underset{x\to 0,y\to 0}{lim}z=\underset{k\to \mathrm{\infty }}{lim}z(M_k)=\underset{k\to \mathrm{\infty }}{lim}\frac{\frac{3}{k}}{\frac{4}{k}-\frac{3}{k}}=3.$$

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

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

Answer: The function can approach any limit.

5. To show that for the function $f(x,y)=\frac{x-y}{x+y}$ there is no limit as $\underset{x\to 0,y\to 0}{lim},$ compute the repeated limits. $$\underset{x\to 0}{lim}(\underset{y\to 0}{lim}f(x,y)),\underset{y\to 0}{lim}(\underset{x\to 0}{lim}f(x,y)).$$

Solution.

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

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

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

Answer: The limit does not exist.

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

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 $k\to \mathrm{\infty }.$ Then $$\underset{x\to 0,y\to 0}{lim}z=\underset{k\to \mathrm{\infty }}{lim}z(M_k)=\underset{k\to \mathrm{\infty }}{lim}\sin \ln (\frac{1}{2k}+\frac{1}{2k})=\underset{k\to \mathrm{\infty }}{lim}\sin \ln \frac{1}{k}.$$

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

Answer: The function does not have a limit.

Homework:

1. $\underset{x\to 0,y\to 0}{lim}\frac{\sin xy}{xy}.$

2 $\underset{x\to 0,y\to 0}{lim}(1+x^2+y^2{)}^{\frac{1}{x^2+y^2}}.$

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

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

Tags: calculus, functions of several variables, iterated limits, limit, mathematical analysis