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Limit of the function x^2-y^2

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     / 2    2\
 lim \x  - y /
x->oo         
$$\lim_{x \to \infty}\left(x^{2} - y^{2}\right)$$
Limit(x^2 - y^2, x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(x^{2} - y^{2}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(x^{2} - y^{2}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{1 - \frac{y^{2}}{x^{2}}}{\frac{1}{x^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{1 - \frac{y^{2}}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{- u^{2} y^{2} + 1}{u^{2}}\right)$$
=
$$\frac{- 0^{2} y^{2} + 1}{0} = \infty$$

The final answer:
$$\lim_{x \to \infty}\left(x^{2} - y^{2}\right) = \infty$$
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
Rapid solution [src]
oo
$$\infty$$
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(x^{2} - y^{2}\right) = \infty$$
$$\lim_{x \to 0^-}\left(x^{2} - y^{2}\right) = - y^{2}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(x^{2} - y^{2}\right) = - y^{2}$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(x^{2} - y^{2}\right) = 1 - y^{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(x^{2} - y^{2}\right) = 1 - y^{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(x^{2} - y^{2}\right) = \infty$$
More at x→-oo