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Limit of the function/ x^2-y^2

Limit of the function x^2-y^2

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     / 2    2\
 lim \x  - y /
x->oo
lim⁡x→∞(x2−y2)\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→∞(x2−y2)\lim_{x \to \infty}\left(x^{2} - y^{2}\right)
Let's divide numerator and denominator by x^2:
lim⁡x→∞(x2−y2)\lim_{x \to \infty}\left(x^{2} - y^{2}\right) =
lim⁡x→∞(1−y2x21x2)\lim_{x \to \infty}\left(\frac{1 - \frac{y^{2}}{x^{2}}}{\frac{1}{x^{2}}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
lim⁡x→∞(1−y2x21x2)=lim⁡u→0+(−u2y2+1u2)\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)
=
−02y2+10=∞\frac{- 0^{2} y^{2} + 1}{0} = \infty

The final answer:
lim⁡x→∞(x2−y2)=∞\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
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→∞(x2−y2)=∞\lim_{x \to \infty}\left(x^{2} - y^{2}\right) = \infty
lim⁡x→0−(x2−y2)=−y2\lim_{x \to 0^-}\left(x^{2} - y^{2}\right) = - y^{2}
More at x→0 from the left
lim⁡x→0+(x2−y2)=−y2\lim_{x \to 0^+}\left(x^{2} - y^{2}\right) = - y^{2}
More at x→0 from the right
lim⁡x→1−(x2−y2)=1−y2\lim_{x \to 1^-}\left(x^{2} - y^{2}\right) = 1 - y^{2}
More at x→1 from the left
lim⁡x→1+(x2−y2)=1−y2\lim_{x \to 1^+}\left(x^{2} - y^{2}\right) = 1 - y^{2}
More at x→1 from the right
lim⁡x→−∞(x2−y2)=∞\lim_{x \to -\infty}\left(x^{2} - y^{2}\right) = \infty
More at x→-oo
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