# Find limit using polar coordinates `x=rcos(t)` ,` y=rsin(t)` , `r=sqrt(x^2+y^2)>= 0` . `lim_(x,y->0,0) yln(x^2+y^2).`

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### 1 Answer

This question has to do with the transformation of Cartesian coordinates to polar coordinates and what it means for this bidirectional limit to exist. We'll start by first substituting the polar coordinates for `x` and `y` using the parameterized functions described in the problem:

`x = rcostheta`

`y = rsintheta`

So:

`lim_(x,y->0,0) yln(x^2+y^2) = lim_(r->0) rsinthetaln(r^2cos^2theta+r^2sin^2theta)`

Notice that the inside of the logarithm can be reduced by using the trig identity `cos^2theta + sin^2theta=1`:

`=lim_(r->0)rsinthetaln(r^2)`

Now, we can try to evaluate the limit. However, notice that as r approaches 0, the log approaches negative infinity and that the r term approaches 0. We'll need another approach:

`=lim_(r->0) sintheta ln(r^2)/(1/r)`

There we go, now the numerator and denominator both approach `+-oo` so we can use calculus students' favorite rule: L' Hopitals' Rule! Keep in mind, `theta` is independent of r (which is good, because in order for a 2-D limit to exist, it must be the same from all directions!).

`=lim_(r->0) sintheta (2r ln(r^2))/(lnr)`

Well, we don't seem to have gone far, have we? Well, we still have a trick with the logs and exponents, so we can set `lnr^2 = 2lnr`:

`=lim_(r->0)sintheta (4rlnr)/(lnr)`

There we go! Now, we can cancel the logarithm terms to end up with an easily calculated limit!

`=lim_(r->0) sintheta*4r=0`

So, in the end, your limit as x and y approach 0 is simply 0.

Hope that helped! ``

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