Multivariable limits: lim (x,y) approaches (1,0) y^2*ln(x)/((x-1)^2+y^2)

[fetante]

Okay can someone help me out with this problem?

lim (x,y) approaches (1,0) y^2*ln(x)/((x-1)^2+y^2).

I have tried to substitute for polar coordinates but have failed.

 

[Deleted member 4993]

Okay can someone help me out with this problem?

lim (x,y) approaches (1,0) y^2*ln(x)/((x-1)^2+y^2).

I have tried to substitute for polar coordinates but have failed.

Please share your work.

 

[fetante]

Please share your work.

lim (x,y) approaches (1,0) y^2*ln(x)/((x-1)^2+y^2).
substitute y = rsinθ, x = 1 + rcosθ.

Now when (x,y)-> (1,0) r->0

lim r->0 (r²sin²(θ)*ln(1+rcosθ)) / ((r²(cos²θ + sin²θ). I know that (cos²θ + sin²θ) = 1

so
. (r²sin²(θ)*ln(1+rcosθ)) / ((r²(1))

simplify
(sin²(θ)*ln(1+rcosθ))

Okay now i´m stuck. Where should i go with this? Am i supposed to use the squeeze theorem maybe?

Edit: I realized i have made some errors, and i corrected them. So

(sin²(θ)*ln(1+rcosθ)) = (sin²(θ)*ln(1+0)) when r->0

So now we have a limited function times a function equal to zero => f = 0.

Am i right?

 

[Deleted member 4993]

lim (x,y) approaches (1,0) y^2*ln(x)/((x-1)^2+y^2).
substitute y = rsinθ, x = 1 + rcosθ.

Now when (x,y)-> (1,0) r->0

lim r->0 (r²sin²(θ)*ln(1+rcosθ)) / ((r²(cos²θ + sin²θ). I know that (cos²θ + sin²θ) = 1

so
. (r²sin²(θ)*ln(1+rcosθ)) / ((r²(1))

simplify
(sin²(θ)*ln(1+rcosθ))

Okay now i´m stuck. Where should i go with this? Am i supposed to use the squeeze theorem maybe?

Edit: I realized i have made some errors, and i corrected them. So

(sin²(θ)*ln(1+rcosθ)) = (sin²(θ)*ln(1+0)) when r->0

So now we have a limited function times a function equal to zero => f = 0.

Am i right?

As far as I can tell, you are correct.