Sometimes, substitution might not be possible. At the same time, there wouldn't be any common factors between the numerator and the denominator.

In such a situation, what we can do is

• Approach the point through different curves.
• The curve should pass through the point

This is generally used to show that the limit does not exist.

If we wish to find $\displaystyle \lim\limits _{( x,y)\rightarrow ( 0,0)} f( x,y)$, we could use $\displaystyle y=mx$ or $\displaystyle y=mx^{2}$. We could use other curves also. However, these are the most commonly used curves.

If we wish to find $\displaystyle \lim _{( x,y)\rightarrow ( 5,3)} f( x,y)$, we could probably use $\displaystyle y=\frac{3}{5} x$ or $\displaystyle y=\frac{3}{25} x^{2}$.

The curve must pass through the point at which we need to find the limit.

$$\boxed{\begin{matrix} \mathbf{If\ we\ get\ different\ limits\ along\ different\ curves,\ we\ can\ say\ that\ the\ limit\ does\ not\ exist\ at\ that\ point.} \end{matrix} }$$

Here, substitution is clearly not possible. There are no common factors ibetween the numerator and the denominator.

Let us approach $\displaystyle ( 0,0)$ along $\displaystyle y=mx$.

$$f( x,y)\Bigl|_{y=mx} =\frac{3x( mx)}{x^{2} +( mx)^{2}} =\frac{3mx^{2}}{x^{2} +m^{2} x^{2}} =\frac{3mx^{2}}{\left( 1+m^{2}\right) x^{2}} =\frac{3m}{1+m^{2}}$$

Now, let us find $\lim\limits_{( x,y)\rightarrow ( 0,0)} f( x,y)\Bigl|_{y=mx}$

$$ \begin{matrix} \lim\limits_{(x,y)\rightarrow (0,0)} f(x,y) |_{y=mx} & = & \lim\limits _{(x,y)\rightarrow (0,0)} \frac{3m}{1+m^2} \\ & & \\ & = & \frac{3m}{1+m^2} \end{matrix} $$

$\mathrm{Here,}$ $\displaystyle \frac{3m}{1+m^{2}}$ $\mathrm{\ is\ a\ constant.\ We\ know\ that\ the\ limit\ of\ a\ constant\ is\ the\ constant\ itself.}$

As we change $m$, we will get different limits.

Here, $m=2$. Therefore,

$$ \begin{matrix} \lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)|_{y=2x} & = & \frac{3(2)}{1+(2)^2} \\ & & \\ & = & \frac{6}{5} \end{matrix} $$

Here, $m=3$. Therefore,

$$ \begin{matrix} \lim\limits_{(x,y)\rightarrow (0,0)} f(x,y)|_{y=3x} & = & \frac{3(3)}{1+(3)^2} \\ & & \\ & = & \frac{9}{10} \end{matrix} $$

What does this suggest?

Approaching $(0,0)$ along different curves gives us different limits. Therefore, the limit of $f(x,y)$ does not exist at $\displaystyle ( 0,0)$.

Again, substituting $\displaystyle ( 0,1)$ into the denominator will give us $\displaystyle 0$. We do not have any common factors between the numerator and the denominator to cancel.

Let us approach $\displaystyle ( 0,1)$ along $\displaystyle y=e^{mx}$. Notice, $\displaystyle e^{mx}$ passes through $\displaystyle ( 0,1)$ for all $\displaystyle m\in ℝ$.

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

$f(x,y)|_{y=e^{mx}}$ is a constant function. However, for differnet values of $m$, we will get different constants.

$$ \lim\limits_{(x,y)\rightarrow (0,1)} f(x,y)|_{y=e^{2x}}= \lim\limits _{(x,y)\rightarrow (0,1)} \frac{2}{1+(2)^2}= \lim\limits _{(x,y)\rightarrow (0,1)}\frac{2}{5} = \frac{2}{5}$$

$$ \lim\limits_{(x,y)\rightarrow (0,1)} f(x,y)|_{y=e^{3x}}= \lim\limits _{(x,y)\rightarrow (0,1)} \frac{3}{1+(3)^2}= \lim\limits _{(x,y)\rightarrow (0,1)} \frac{3}{10} = \frac{3}{10} $$

The limit of a constant function is the constant itself. These constants are different for diffent $m$. Therefore, the limit of $f(x,y)$ at $(0,1)$ does not exist.

There could be some functions for which both substitution and cancelling the common factors might not be possible. At the same time, the limit along different curves will be equal. However, this is not enough to conclude that the limit exists. We will see one such case now.

Substituting $\displaystyle ( 0,0)$ into the denominator will not be possible. The denominator will become $\displaystyle 0$. There is no common factor between the numerator and the denominator.

Let us try to approach $\displaystyle ( 0,0)$ along different curves.

$$ \begin{matrix} f(x,y)|_{y=mx} & = & \frac{x^{5} (mx)}{x^{2} +(mx)^{2}} & = & \frac{mx^6}{x^2(1+m^2)} & = & \frac{mx^4}{1+m^2} \end{matrix} $$

Let us find the limit of $f(x,y)$ at $(0,0)$ along $y=mx$

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx} & = & \lim\limits _ {(x,y)\rightarrow (0,0)}\frac{mx^4}{1+m^2} & = & \frac{m(0)^4}{1+m^2} & = & 0 \end{matrix} $$

Irrespective of the value of $m$, the limit of $f(x,y)$ at $(0,0)$ along $y=mx$ is $0$

$$ \begin{matrix} f(x,y)|_{y=mx^2} & = & \frac{x^{5} (mx^2)}{x^{2} +(mx^2)^{2}} & = & \frac{mx^7}{x^2(1+m^2x^2)} & = & \frac{mx^5}{1+m^2x^2} \end{matrix} $$

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

$$ \begin{matrix} \lim\limits _ {(x,y)\rightarrow (0,0)} f(x,y)|_{y=mx^2} & = & \lim\limits _{(x,y)\rightarrow (0,0)}\frac{mx^5}{1+m^2x^2} & = & \frac{m(0)^5}{1+m^2(0)^2} & = & 0 \end{matrix} $$

Again, irrespective of the value of $m$, the limit of $f(x,y)$ at $(0,0)$ along $y=mx^2$ is $0$

From these observations, can we conclude that the limit of $f(x,y)$ at $(0,0)$ is $0$?

NO. To make this conclusion, as we saw earlier, we have to see if the limit of $f(x,y)$ at $(0,0)$ is $0$ along all possible curves passing through $(0,0)$.

This is difficult to check. We can come up with as many curves as we want.

Let us revisit this function after looking at The Sandwich Theorem.