OPEN BALL CENTERED AT A POINT

Let $x\in ℝ^d$. Then, an open ball of radius $r>0$ centered at $x$ is given by the following set $:$

$$ \lbrace v\in ℝ^d:‖v-x‖ < r \rbrace $$

This set is the set of all the points in $ℝ^d$, whose distance from $(x)$ is less than $r$, where $r>0$.

Let us look at an open ball of radius $3$ centered at the origin in $ℝ^2$ and $ℝ^3$

THE SANDWICH THEOREM

(Taken from Exercise 14.2, Thomas' Calculus, ${14^{th}}$ edition, and rephrased the definition using the notions and words used in our course)

The sandwich theorem for functions of two variables states that if

$$ g(x,y)⩽f(x,y)⩽h(x,y)$$

for all $(x,y)$ in an open ball of radius $r>0$ centered at $(x_{0} ,y_{0})$, except possibly at $(x_0,y_0)$ and if $g$ and $\displaystyle h$ have the same finite limit $\displaystyle L$ as $\displaystyle ( x,y)\rightarrow ( x_{0} ,y_{0})$, then

$$\lim\limits _ {(x,y)\rightarrow (x_{0},y_{0})} f(x,y) =L$$

In simple words, if $:-$

• $g(x,y)⩽f(x,y)⩽h(x,y)$ is true for all $(x,y)$ in an open ball of radius $r>0$ centered at $\displaystyle ( x_{0} ,y_{0})$, except possibly at $(x_0,y_0)$.

• $\lim\limits _ {(x,y)\rightarrow (x_{0},y_{0})} g(x,y) = L \mathrm{\ \ and } \lim\limits _ {(x,y)\rightarrow (x_{0},y_{0})} g(x,y) = L $

then,

$$\lim\limits _ {(x,y)\rightarrow (x_{0},y_{0})} f(x,y) =L$$

FEW COMMON BOUNDS

• $0⩽x^2⩽x^2+y^2$

• $-1⩽x\sin(1/x)⩽1$

• $-x⩽x\sin(1/x)⩽x$

• $0⩽2|xy|⩽x^2+y^2$

EXAMPLE-3 (Continuation)

Let us continue from where we stopped in EXAMPLE-3

$\mathrm{The\ function\ was\ } f(x,y)=\frac{x^5y}{x^2+y^2}.\mathrm{\ We\ wanted\ to\ find\ } \lim _{( x,y)\rightarrow ( 0,0)} f( x,y).$

We observed that the limits along the curves of the form $y=mx$ and $y=mx^2$ were equal to $0$. However, we stated the reason why this observation is not not enough to conclude that the limit of $f(x,y)$ along $(0,0)$ exists.

Let use observe the following bound.

$$ \begin{matrix} 0⩽x^2⩽x^2+y^2 \ \ 0⩽\frac{x^2}{x^2+y^2}⩽1 \ \ 0⩽\frac{x^5y}{x^2+y^2}⩽x^3y \ \ 0⩽f(x,y)⩽x^3y \end{matrix} $$

$\mathrm{Notice} :- \ \lim\limits _{( x,y)\rightarrow ( 0,0)}( 0) =0\ \ \mathrm{and} \ \lim\limits _{( x,y)\rightarrow ( 0,0)}\left( x^{3} y\right) =0.$

Using sandwich theorem, we can say that $:$-

$$ \lim\limits _{( x,y)\rightarrow ( 0,0)} f( x,y) =0 $$

ACTIVITY 9.4 (QUESTION-10)

Consider the function $f:ℝ^2\rightarrow ℝ$ defined as $:$

$$ \begin{matrix} f(x,y)=\frac{x^ky}{x^{2n}+y^{2n}} & \mathrm{if\ }x,y\ne 0 \ & \ f(x,y)=0 & \mathrm{if\ }x=y=0 \end{matrix} $$

where $k,n\in ℕ \backslash \lbrace 0 \rbrace $. Which of the following is(are) true about $f$

1. If $k=2n-1$, then the limit at $\displaystyle ( 0,0)$ exists and is equal to 0.

2. If $k<2n-1$, then the limit at $(0,0)$ does not exist.

3. If $k>2n$, then the limit at $(0,0)$ always exists and is equal to 0.

4. If $k>2n$, then the limit at $\displaystyle ( 0,0)$ does not exist.

We can write $\frac{x^ky}{x^{2n}+y^{2n}}$ as follows $:$

$$ \begin{matrix} \frac{x^ky}{x^{2n}+y^{2n}} & = & \frac{x^ky/x^{2n}}{(x^{2n}+y^{2n})/x^{2n}} \ & & \ & = & \frac{x^{k-2n}y}{1+(y/x)^{2n}} \end{matrix} $$

We shall use this for all the cases below.

Let us approach $(0,0$ along $y=mx$

$$ \begin{matrix} f(x,y)|_{y=mx} & = & \frac{x^{k-2n}(mx)}{1+(mx/x)^{2n}} & = & \frac{m(x)^{k-2n+1}}{1+(m)^{2n}} \end{matrix} $$

CASE-1 ($k=2n+1$)

If $k=2n-1$, the $k-2n+1=0$. Therefore, $x^{k-2n+1}=x^0=1$.

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx} & = & \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{m}{1+m^{2n}} \end{matrix} $$

Since $\frac{m}{1+m^{2n}}$ depends on $m$, we can say that $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y) $ does not exist.

Hence, option 1 is incorrect.

CASE-2 ($k<2n-1$)

If $k<2n-1$, then $k-2n+1<0$. As a result $:$

$$ \begin{matrix} \frac{m(x)^{k-2n+1}}{1+(m)^{2n}} & = & \frac{m}{(x^{2n-k-1})(1+(m)^{2n})} \end{matrix} $$

Here, $2n-k-1>0$.

Notice $:$ $\frac{m}{1+m^{2n}}$ is a constant in $\frac{m}{(x^{2n-k-1})(1+(m)^{2n})}$

Let us try to find $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx}$.

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx} & = & \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{m}{(x^{2n-k-1})(1+(m)^{2n})} & = & (\frac{m}{1+m^{2n}}) \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{1}{x^{2n-k-1}} \end{matrix} $$

$\mathrm{We\ know\ that\ }\lim\limits _ {x\rightarrow 0} \frac{1}{x^d} \mathrm{\ does\ not\ exist\ (in\ the\ finite\ sense).\ } \mathrm{Therefore,\ } \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{1}{x^{2n-k-1}} \mathrm{\ does\ not\ exist.\ } \mathrm{From\ this\ we\ can\ say\ that}\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx} \mathrm{\ does\ not\ exist.} $

Hence, option 2 is correct.

CASE-3 ($k>2n$)

If $k>2n$, then $k-2n>0$. Therefore, $k-2n+1>0$

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{m(x)^{k-2n+1}}{1+(m)^{2n}} & = & \lim\limits _ {(x,y)\rightarrow (0,0)} \frac{m(0)^{k-2n+1}}{1+(m)^{2n}} & = & 0 \end{matrix} $$

$\mathrm{We\ can\ see\ that\ } \lim\limits _ {(x,y) \rightarrow (0,0)} f(x,y)| _ {y=mx} = 0 \mathrm{,\ irrespective\ of\ the\ value\ of\ } m.$

Let us try to approach $(0,0)$ along $y=mx^2$

$$ \begin{matrix} f(x,y)|_{y=mx^2} & = & \frac{x^{k-2n}(mx^2)}{1+(mx^2/x)^{2n}} & = & \frac{m(x)^{k-2n+2}}{1+(mx)^{2n}} \end{matrix} $$

Since $k>2n$, we can say that $k-2n+2>0$

Let us try to find the limit of $f(x,y)$ at $(0,0)$ along the curve $y=mx^2$.

$$ \begin{matrix} \lim\limits _ {(x,y) \rightarrow (0,0)} f(x,y)|_{y=mx^2} & = & \lim\limits _ {(x,y) \rightarrow (0,0)} \frac{m(x)^{k-2n+2}}{1+(mx)^{2n}} & = & \frac{m(0)^{k-2n+2}}{1+(m(0))^{2n}} & = & 0 \end{matrix} $$

$\mathrm{Again,\ we\ can\ see\ that\ } \lim\limits _ {(x,y) \rightarrow (0,0)} f(x,y)| _ {y=mx^2} = 0 \mathrm{,\ irrespective\ of\ the\ value\ of\ } m.$

Let us see if we can apply sandwich theorem here.

$$ \begin{matrix} 0⩽x^{2n}⩽x^{2n}+y^{2n} \ \ 0⩽\frac{x^{2n}}{x^{2n}+y^{2n}}⩽1 \ \end{matrix} $$

Let us try to multiply this inequality with $x^{k-2n}y$. We need to consider the fact that $x^{k-2n}y$ can be positive or negative or equal to $0$. Therefore, we will evaluate both the cases separately.

Note $:$ $x^{k-2n}y$ will be equal to zero if either $x$ or $y$ is equal to $0$

Since $k>2n$, we can substitute $(0,0)$ in place of $(x,y)$ in $x^{k-2n}y.$ There is no issue of zero being in the denominator.

If $x^{k-2n}y>0$

$$ \begin{matrix} 0⩽\frac{x^{2n}(x^{k-2n}y)}{x^{2n}+y^{2n}}⩽x^{k-2n}y \
\ 0⩽\frac{x^ky}{x^{2n}+y^{2n}}⩽x^{k-2n}y \ \ 0⩽f(x,y)⩽x^{k-2n}y \end{matrix} $$

Now, $\lim\limits _ {(x,y)\rightarrow (0,0)} 0=0 \mathrm{\ and\ } \lim\limits _ {(x,y)\rightarrow (0,0)} x^{k-2n}y=0$

By sandwich theorem, $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)=0$

If $x^{k-2n}y<0$

$$ \begin{matrix} 0⩽\frac{x^{2n}(x^{k-2n}y)}{x^{2n}+y^{2n}}⩽x^{k-2n}y \
\ x^{k-2n}y⩽\frac{x^ky}{x^{2n}+y^{2n}}⩽0 \ \ x^{k-2n}y⩽f(x,y)⩽0 \end{matrix} $$

Now, $\lim\limits _ {(x,y)\rightarrow (0,0)} x^{k-2n}y=0 \mathrm{\ and\ } \lim\limits _ {(x,y)\rightarrow (0,0)} 0=0$

By sandwich theorem, $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)=0$

If $x^{k-2n}y=0$

This is possible if $x=0$ or $y=0$.

If $y=0$, we are looking at the case of $y=mx$. We have already seen that $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)| {y=mx}=0$. From this we can see that $\lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=0} = 0$.

Let us look at the case where $x=0$

$$ \begin{matrix} f(x,y)|_{x=0} & = & \frac{0^ky}{0^{2m}+y^{2m}} & = & 0 \end{matrix} $$

$\mathrm{Let\ us\ find} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{x=0}$

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{x=0} & = & \lim\limits _ {(x,y)\rightarrow (0,0)} 0 & = & 0 \end{matrix} $$

We have looked at all the possibilities. From these observations, we can conclude that the limit of $f(x,y)$ at $(0,0)$ exists and is equal to $0$

Hence, option 3 is correct and option 4 is incorrect.