# How do I find the limit of (xy)/sqrt (x^2+y^2)?

##### 2 Answers
Feb 18, 2015

I suppose that the limit is:

${\lim}_{\left(x , y\right) \rightarrow \left(0 , 0\right)} \frac{x y}{\sqrt{{x}^{2} + {y}^{2}}}$.

The answer is: $0$.

This limit is in the indecision form: $\frac{0}{0}$.

The limits of a function of more than one variable are really different from the ones of one variable. The variable $x$ can "go" to its limit (e.g. $a$) "only" in two directions (${a}^{+} , {a}^{-}$). In more than one variable the directions are infinite! So, if we want to demonstrate that one limit doesn't exist it is simply, because we can choose two different directions that drive to different values.

The way to demonstrate, and to calculate, a limit making only one limit is to make a calculation for every direction.
It seems to be not so easy...

but

if we change the coordinate system from cartesian to polar, we will have two new variables $\rho , \theta$ and $\theta$ will give us all the directions in one shot!

If the limit will depend only from $\rho$ the limit exist!

Now let's do the exercise.

I remember that:

$x = \rho \cos \theta$
$y = \rho \sin \theta$

and if

$\left(x , y\right) \rightarrow \left(0 , 0\right)$

than

$\rho \rightarrow 0$.

So:

${\lim}_{\rho \rightarrow 0} \frac{\rho \cos \theta \rho \sin \theta}{\sqrt{{\rho}^{2} {\cos}^{2} \theta + {\rho}^{2} {\sin}^{2} \theta}} =$

${\lim}_{\rho \rightarrow 0} \frac{{\rho}^{2} \cos \theta \sin \theta}{\sqrt{{\rho}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)}} =$

${\lim}_{\rho \rightarrow 0} \frac{{\rho}^{2} \cos \theta \sin \theta}{\rho \sqrt{\left({\cos}^{2} \theta + {\sin}^{2} \theta\right)}} =$

${\lim}_{\rho \rightarrow 0} \frac{{\rho}^{2} \cos \theta \sin \theta}{\rho} =$

${\lim}_{\rho \rightarrow 0} \rho \cos \theta \sin \theta = 0$

and, it's easy to say, this limit doesn't depends from $\theta$!

Dec 18, 2015

You can tell the limit is going to be zero by comparing the degrees of the top and bottom.

#### Explanation:

The numerator has "2nd order smallness" - it's "tiny" - because of the $x \cdot y$.

The denominator has only "1st order of smallness" - it's just "small" - because the $\sqrt{{x}^{2} + {y}^{2}}$ is around the same size as the x or y.

So the quotient will be "tiny/small" which goes to 0.

"Number sense" - If x and y are around 1/100, the top will be about 1/10000, and $\sqrt{{x}^{2} + {y}^{2}}$ will be about 1/70, so the quotient is about 7/1000, close to zero.