Question

How do I find the limit of $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$?

Answer 1

This question is from the topic of limits. We will first find the value just by putting the limits putting x and y as zero. From here, we will get indeterminate or undefined form. So, we will use different methods to solve this question. We will convert the x and y coordinates into polar coordinates or we can say we will convert the x and y in the form of $r$ and $\theta$ to solve this question correctly.

Complete step by step answer:
Let us solve this question.
In this question, we have to find the limit $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ or we can say we have to find the limit of the term $\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ having limits $(x,y)\to (0,0)$.
Putting the limits in the term$\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$, we will get
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\dfrac{0\times 0}{0+0}=\dfrac{0}{0}$
The term $\dfrac{0}{0}$ is indeterminate form or we can say that this term is undefined.
So, we will do this question using a different way.
We will use polar coordinates for solving this question in a better way.
As we know that x and y coordinates can be written in the form of polar coordinates.
$x=r\cos \theta$
$y=r\sin \theta$
The terms $r$ and $\theta$ are polar coordinates. As x and y tends to 0, then r will tend to be zero.
So, after putting the values of x and y in the form of polar coordinates in the limit, we can write
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{r\cos \theta \times r\sin \theta }{\sqrt{{{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}}}$
As we have taken limits$(x,y)\to (0,0)$, then we can write the limit as $r\to 0$
The above equation can also be written as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{r\cos \theta \times r\sin \theta }{\sqrt{{{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}}}$
The above equation can also be written as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{{{r}^{2}}\cos \theta \times \sin \theta }{\sqrt{{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta }}$
The above equation can also be written as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{{{r}^{2}}\cos \theta \times \sin \theta }{\sqrt{{{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}}$
As we know the identity that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and $\sqrt{{{r}^{2}}}$ will be equal to$r$.
So, we can write the above equation as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{{{r}^{2}}\cos \theta \times \sin \theta }{\sqrt{{{r}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{{{r}^{2}}\cos \theta \times \sin \theta }{r}$
The above equation can also be written as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}r\cos \theta \times \sin \theta$
Now, we will put the value of limit, that is we will write r=0. We get
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\left( 0\times \cos \theta \times \sin \theta \right)=0$
As the value of sin and cos is always between 0 and 1, so they will be finite and not infinite.
Hence, we get that the value of $\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$as 0.

Note:
We should have a better knowledge in the topic of limits to solve this type of question easily. Don’t forget the trigonometric identity like: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$.
We can solve this question by an alternate method.
We have found above that
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}r\cos \theta \times \sin \theta$
The above equation can be written as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\left( \dfrac{r}{2}\times 2\times \cos \theta \times \sin \theta \right)$
As we know that $2\times \cos \theta \times \sin \theta =\sin 2\theta$, so we can write the above equation as
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta$
Using squeeze theorem, we can write
$-1\le \sin 2\theta \le 1$
After multiplying $\dfrac{r}{2}$ in the above equation, we get
$-\dfrac{r}{2}\le \dfrac{r}{2}\sin 2\theta \le \dfrac{r}{2}$
Now, putting the limits in the above equation, we get
$\displaystyle \lim_{r\to 0}-\dfrac{r}{2}\le \displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta \le \displaystyle \lim_{r\to 0}\dfrac{r}{2}$
Hence, after putting the value of r in all the terms, we get
$0\le \displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta \le 0$
So, we can say that the value of $\displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta$ is 0.
Or, we can write
$\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta =0$