\lim_(x,y \to \infty) (x+y)/(x^(2) +y^(2)-xy )

Question

`\lim_(x,y \to \infty) (x+y)/(x^(2) +y^(2)-xy )`

Answer 1

To evaluate the limit
`\sf \lim_(x,y \to \infty) (x+y)/(x^(2) +y^(2)-xy) \\`, we can analyze the behavior of the expression as both
`\sf x \\` and
`\sf y \\` approach infinity.

Let's consider the numerator
`\sf x + y \\` and the denominator
`\sf x^(2) + y^(2) - xy \\` separately.

For the numerator, as both
`\sf x \\` and
`\sf y \\` approach infinity, their sum
`\sf x+y \\` will also approach infinity.

For the denominator, we can rewrite it as
`\sf (x-y)^2 + 2xy \\`. As
`\sf x` and
`\sf y` approach infinity, the terms
`\sf (x-y)^2 \\` and
`\sf 2xy \\` will also approach infinity. Therefore, the denominator will also approach infinity.

Now, let's consider the entire fraction
`\sf (x+y)/(x^(2) +y^(2)-xy) \\`. Since both the numerator and denominator approach infinity, we have an indeterminate form of
`\sf (\infty)/(\infty) \\`.

To evaluate this indeterminate form, we can apply techniques such as L'Hôpital's rule or algebraic manipulations. However, in this case, we can simplify the expression further.

By dividing both the numerator and denominator by
`\sf x^(2) \\`, we get:

`\sf \lim_(x,y \to \infty) ((x)/(x^(2)) + (y)/(x^(2)))/(1 + (y^(2))/(x^(2)) - (xy)/(x^(2))) \\`

As
`\sf x` approaches infinity, the terms
`\sf (x)/(x^(2)) \\` and
`\sf (y)/(x^(2)) \\` both approach zero. Similarly, the term
`\sf (y^(2))/(x^(2))` and
`\sf (xy)/(x^(2)) \\` also approach zero.

Therefore, the limit simplifies to:

`\sf \lim_(x,y \to \infty) (0 + 0)/(1 + 0 - 0) = (0)/(1) = 0 \\`

Hence, the limit
`\sf \lim_(x,y \to \infty) (x+y)/(x^(2) +y^(2)-xy) \\` is equal to 0.