Question

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

Answer 1

Hint: 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
\[\displaystyle \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}=\displaystyle \lim_{r\to 0}\dfrac{r}{2}\sin 2\theta =0\]