Question

Show sin(x^2 y^2)/(x^2 y^2) is cont.

Answer 1

The function
`\( \frac{{\sin(x^2 y^2)}}{{x^2 y^2}} \)` is continuous everywhere except at the point where x = 0 and y = 0.

Step-by-step explanation:

To assess the continuity of the function
`\( \frac{{\sin(x^2 y^2)}}{{x^2 y^2}} \)`, we need to examine its behavior around the point (0,0). The function is composed of two parts:
`\( \sin(x^2 y^2) \) and \( x^2 y^2 \)` in the denominator. As x and y approach 0,
`\( x^2 y^2 \)`the denominator becomes 0, making the function undefined at that point. However,
`\( \sin(x^2 y^2) \)` approaches 0 as x and y approach 0, which implies that the function is continuous everywhere except at x = 0 and y = 0.

The function exhibits continuity in its behavior throughout its domain, except at the origin where x = 0 and y = 0. At this specific point, the function encounters a singularity due to the indeterminate form of
`\( (0)/(0) \)`, signifying a lack of continuity. Nonetheless, away from this point, the function
`\( \frac{{\sin(x^2 y^2)}}{{x^2 y^2}} \)` behaves smoothly and continuously across its domain, with no disruptions or discontinuities. Understanding the points of discontinuity helps in defining the regions where the function remains continuous and assists in analyzing its behavior within its defined domain.