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Use the method of your choice to evaluate the following limits. $$\lim _{(x, y) \rightarrow(1,1)} \frac{x^{2}+x y-2 y^{2}}{2 x^{2}-x y-y^{2}}$$

Short Answer

Expert verified
Answer: The limit of the given expression as (x, y) approaches (1,1) is 1.

Step by step solution

01

Factorization

First, we need to factorize the numerator and the denominator to see if there are any common factors that can be cancelled out. The numerator expression is \(x^2 + xy - 2y^2\), which can be factored as \((x-1)(x+2y)\). The denominator expression is \(2x^2 - xy - y^2\), which can be factored as \((x-1)(2x+y)\). Now, we can rewrite our original expression as a simplified fraction: $$\frac{(x-1)(x+2y)}{(x-1)(2x+y)}$$
02

Cancelling Common Terms

We see that both the numerator and the denominator have the common factor \((x-1)\). So, we can cancel it out: $$\frac{(x-1)(x+2y)}{(x-1)(2x+y)} = \frac{x+2y}{2x+y}$$
03

Evaluate the Limit

Now that the expression is simplified, we can evaluate the limit as \((x, y) \to (1, 1)\): $$\lim_{(x, y) \rightarrow(1,1)}\frac{x + 2y}{2x + y}$$ Substitute the values \((x, y) = (1, 1)\) into the simplified expression: $$\frac{1 + 2(1)}{2(1) + 1}$$ Evaluate the expression: $$\frac{3}{3}$$
04

Conclusion

In conclusion, the limit of the given expression as \((x, y)\) approaches \((1,1)\) is equal to 1: $$\lim_{(x, y) \rightarrow(1,1)} \frac{x^{2}+x y-2 y^{2}}{2 x^{2}-x y-y^{2}} = 1$$

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