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Find the limit. lim (x,y)→(1,29)X/√x+yDiscuss the continuity of the function. continuous for x+y≥0 continuous for x+y=0 continuous for x+y>0 continuous everywhere except (0,0) continuous everywhere

User Tao Nhu
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Final Answer:

The limit as (x, y) approaches (1, 29) of x/√(x+y) is undefined. The function is continuous everywhere except at the point (0, 0).

Step-by-step explanation:

The given function is f(x, y) = x/√(x+y). To find the limit as (x, y) approaches (1, 29), substitute these values into the function:

lim (x,y)→(1,29) x/√(x+y)

= 1/√(1+29)

= 1/√30

Since the denominator is nonzero, the limit exists and is equal to 1/√30. However, the limit is undefined at (0, 0) as the denominator becomes zero, indicating a potential discontinuity at this point.

Now, let's discuss the continuity of the function. The function is continuous everywhere except at points where the denominator becomes zero, which is x+y=0. This implies that the function is continuous for x+y≠0.

Therefore, the function is continuous everywhere except at the point (0, 0). In other words, the function is continuous for x+y>0, making it undefined for x+y=0 and continuous for x+y<0. This analysis is crucial for understanding the behavior of the function across different regions of the xy-plane.

User Shahjad Ans Sunny
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