Question

Find the following limit. \(\lim _{(x, y) \rightarrow(0,1)} \tan ^{-1}\left[\frac{x^{2}+1}{x^{2}+(y-1)^{2}}\right]\)

Short answer

The limit as (x, y) approaches (0, 1) for the function is \( \frac{\pi}{4} \).

Step-by-step solution

Substitute the values of x and y

Given that the limit as (x, y) approaches (0, 1), start by substituting x = 0 and y = 1 into the fraction inside the inverse tangent function, we get \( \frac{{0^{2}+1}}{{0^{2} + (1-1)^{2}} \} = 1. Remember that we're yet to take the limit of this quotient.

Solve the limit of the arctan function

Since we have the arctangent of 1 and the limit as x approaches 0 is the same as y approaches 1, we now need to take the limit of the arctan (1). The limit as x approaches any real number a of any constant function, including trigonometric functions, is the function's value at a. Thus, the limit of arctan (1) as x approaches 0 is arctan (1), which is \( \frac{\pi}{4} \).

Give the final answer

Having gone through all other steps, the answer to the limit is the result from Step 2. This is the final step of the problem.