Question

The function f(x, y)=\(\frac{2xy-3x}{3x-2y}\) is continuous at (1, 1).

(a) True

(b) False

I have been asked this question in semester exam.

Query is from Continuity in division Complex Function Theory of Engineering Mathematics

Answer 1

Correct answer is (b) False

Explanation: Case 1: \(\lim_{x \to 1 \\ y \to 1}\frac{2xy-3x}{3x-2y}\)

\(\lim_{y \to 1}\frac{2y-3}{3-2y}\)=\(\frac{-1}{1}\)=-1

Case 2: \(\lim_{y \to 1 \\ x \to 1}\frac{2xy-3x}{3x-2y}\)

\(\lim_{x \to 1}\frac{2x-3x}{3x-2}\)=\(\frac{-1}{1}\)=-1

Case 3: \(\lim_{y \to mx \\ x \to 0}\frac{2xy-3x}{3x-2y}\)

\(\lim_{x \to 0}\frac{2mx^2-3x}{3x-2mx}\)=\(\frac{-3}{3-2m}\)

Since the cases are not same, f(x, y) is discontinuous at x=1.