The following problems illustrate different techniques used to evaluate limits of multivariable functions,
including direct substitution, algebraic simplification, polar coordinates, and the path method.
Question 2: Discuss the Continuity of the following functions at given points.
Solved Problems with Collapsible Solutions
To check continuity at (0,0), we verify whether:
$$
\lim_{(x,y)\to(0,0)} f(x,y) = f(0,0)
$$
If the limit does not exist or is not equal to the function value,
then the function is not continuous at (0,0).
Problem I
$$
f(x,y)=
\begin{cases}
\dfrac{2x^2y}{x^4+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx^2$:
Numerator:
$$
2x^2(mx^2)=2mx^4
$$
Denominator:
$$
x^4+m^2x^4=x^4(1+m^2)
$$
Thus,
$$
f(x,y)=\frac{2m}{1+m^2}
$$
This depends on $m$.
Example:
- If $m=0$, limit = 0
- If $m=1$, limit = 1
Since limit depends on path, it does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem II
$$
f(x,y)=
\begin{cases}
\dfrac{2xy}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx$:
$$
f(x,y)=\frac{2mx^2}{x^2(1+m^2)}
=\frac{2m}{1+m^2}
$$
This depends on $m$.
Therefore, the limit does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem III
$$
f(x,y)=
\begin{cases}
\dfrac{x^2-y^2}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Take the path $y=mx$:
$$
\frac{x^2-m^2x^2}{x^2+m^2x^2}
=
\frac{1-m^2}{1+m^2}
$$
Different values for different $m$:
- If $m=0$, limit = 1
- If $m=1$, limit = 0
Hence limit does not exist.
$$
\boxed{\text{Not Continuous at } (0,0)}
$$
Problem IV
$$
f(x,y)=
\begin{cases}
\cos\!\left(\dfrac{x^3-y^3}{x^2+y^2}\right), & (x,y)\ne(0,0) \\
1, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Use polar coordinates:
$$
x=r\cos\theta, \quad y=r\sin\theta
$$
Then,
$$
\frac{x^3-y^3}{x^2+y^2}
=
r(\cos^3\theta-\sin^3\theta)
$$
As $r\to0$:
$$
r(\cos^3\theta-\sin^3\theta)\to0
$$
Thus,
$$
\cos(0)=1
$$
Since limit equals function value:
$$
\boxed{\text{Continuous at } (0,0)}
$$
Problem V
$$
f(x,y)=
\begin{cases}
\dfrac{x^3y}{x^2+y^2}, & (x,y)\ne(0,0) \\
0, & (x,y)=(0,0)
\end{cases}
$$
Click to View Solution
Use polar coordinates:
$$
x=r\cos\theta, \quad y=r\sin\theta
$$
Then,
$$
\frac{r^3\cos^3\theta \cdot r\sin\theta}{r^2}
=
r^2\cos^3\theta\sin\theta
$$
As $r\to0$:
$$
\to0
$$
Since limit equals $f(0,0)=0$:
$$
\boxed{\text{Continuous at } (0,0)}
$$
