Question

EXAMPLE 3 If f(x, y) = 4xy2 7x2 + y4 , does lim (x, y)→(0, 0) f(x, y) exist? SOLUTION Let's try to save some time by letting (x, y) → (0, 0) along any nonvertical line through the origin. Then y = mx, where m is the slope, and f(x, y) = f(x, mx) = 4x 2 7x2 + (mx)4 = 7x2 + m4x4 = 7 + m4x2 .

Answer 1

Answer:

Limit of the function exists.

Step-by-step explanation:

Given the function f(x,y) = $\frac{4xy^{2} }{7x^{2} + y^{4} }$, we are to show that lim (x, y)→(0, 0) f(x, y) exist. To show that, the following steps must be followed.

$\lim_{(x,y) \to (0,0)} \frac{4xy^{2} }{7x^{2} + y^{4} }$

substituting the limit x = 0 and y = 0 into the function we have;

$\frac{4(0)^{2} }{7(0)^{2} + (0)^{4} }\\= \frac{0}{0} (indeterminate)$

Since we got an indeterminate function, we will then substitute y = mx into the function as shown;

$\lim_{(x,mx) \to (0,0)} \frac{4x(mx)^{2} }{7x^{2} + (mx)^{4} }\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2} x^{3} }{7x^{2} + m^{4}x^{4} }\\\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2} x^{3} }{x^{2}(7 + m^{4} x^{2}) }\\\lim_{(x,mx) \to (0,0)} \frac{4m^{2}x }{7 + m^{4} x^{2} }$

Substituting x = 0 , the limit of the function becomes;

$\frac{4m^{2}(0) }{7 + m^{4} (0)^{2} }\\= \frac{0}{7}\\ = 0$

Since the limit of the function gives a finite value of 0 (the limit tends to 0). This shows that the limit exists.