Question

Find \(lt_{(x,y)\rightarrow(0,0)}\frac{x^3+3xy^2-xy^2}{x^2+xy}\)

(a) 0

(b) ∞

(c) 1

(d) -1

I got this question in homework.

My question is taken from Limits and Derivatives of Several Variables topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

The correct choice is (a) 0

Best explanation: Converting into Polar form we have

=\(lt_{r\rightarrow 0}\frac{r^3.cos^3(\theta))+3(r^2.cos^2(\theta))(r.sin(\theta))-(r.cos(\theta))(r^2.sin^2(\theta))}{(r^2cos^2(\theta))+r^2sin(\theta)cos(\theta)}\)

=\(lt_{r\rightarrow 0}\frac{r^3}{r^2}\times (\frac{cos^3(\theta)+3(cos^2(\theta))(sin(\theta))-(cos(\theta))(sin^2(\theta))}{(cos^2(\theta))+sin(\theta)cos(\theta)})\)

=\(lt_{r\rightarrow 0}(r)\times (\frac{cos^3(\theta)+3(cos^2(\theta))(sin(\theta))-(cos(\theta))(sin^2(\theta))}{(cos^2(\theta))+sin(\theta)cos(\theta)})\)

=0