Use polar coordinates to find the limit. [If are (r, theta) polar coordinates of the point (x,y) with r&gt;=0, note that as (x,y) tends to (0,0).] lim(x,y) tends to (0,0)(x2+y2)ln⁡(x2+y2)

Substitute into polar coordinates, that is:x=rcos⁡0y=rsin⁡0Thereforelim(x,y)⇒(0,0)(x2+y2)ln⁡(x2+y2)=limr⇒0+(r2cos20+r2sin20)ln⁡(r2cos20+r2sin20)=limr⇒0+r2(cos20+sin20)ln⁡(r2(cos20+sin20))=limr⇒0+r2ln⁡r2=limr⇒0+ln⁡r2r−2now we have −∞∞ type of limit, and we can use L'Hopital rule:=limr⇒0+∂∂r(ln⁡r2)∂∂r(r−2)=limr⇒0+1r2⋅2r−2r−3=limr⇒0+r4−2r2=limr⇒0+r2−2=0Result:lim(x,y)⇒(0,0)(x2+y2)ln⁡(x2+y2)=0

Expert Answer to "Use polar coordinates to find the limit. [If..."

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

Miguel Mathis

Open question

2022-08-18

Use polar coordinates to find the limit. [If are (r, theta) polar coordinates of the point (x,y) with r>=0, note that as (x,y) tends to (0,0).]

lim (x, y)

tends to (0,0)

(x^{2} + y^{2}) \ln (x^{2} + y^{2})

Answer & Explanation

Karsyn Lam

Beginner2022-08-19Added 7 answers

Substitute into polar coordinates, that is:

x = r \cos 0

y = r \sin 0

Therefore

lim_{(x, y) \Rightarrow (0, 0)} (x^{2} + y^{2}) \ln (x^{2} + y^{2}) = lim_{r \Rightarrow 0^{+}} (r^{2} \cos^{2} 0 + r^{2} \sin^{2} 0) \ln (r^{2} \cos^{2} 0 + r^{2} \sin^{2} 0)

= lim_{r \Rightarrow 0^{+}} r^{2} (\cos^{2} 0 + \sin^{2} 0) \ln (r^{2} (\cos^{2} 0 + \sin^{2} 0))

= lim_{r \Rightarrow 0^{+}} r^{2} {\ln r}^{2}

= lim_{r \Rightarrow 0^{+}} \frac{{\ln r}^{2}}{r^{- 2}}

now we have

\frac{- \infty}{\infty}

type of limit, and we can use L'Hopital rule:

= lim_{r \Rightarrow 0^{+}} \frac{\frac{\partial}{\partial r} ({\ln r}^{2})}{\frac{\partial}{\partial r} (r^{- 2})}

= lim_{r \Rightarrow 0^{+}} \frac{\frac{1}{r^{2}} \cdot 2 r}{- 2 r^{- 3}}

= lim_{r \Rightarrow 0^{+}} \frac{r^{4}}{- 2 r^{2}}

= lim_{r \Rightarrow 0^{+}} \frac{r^{2}}{- 2}

=0
Result:

lim_{(x, y) \Rightarrow (0, 0)} (x^{2} + y^{2}) \ln (x^{2} + y^{2}) = 0

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