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Limit of the function x^2+x*y

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     / 2      \
 lim \x  + x*y/
x->oo          
limx(x2+xy)\lim_{x \to \infty}\left(x^{2} + x y\right)
Limit(x^2 + x*y, x, oo, dir='-')
Detail solution
Let's take the limit
limx(x2+xy)\lim_{x \to \infty}\left(x^{2} + x y\right)
Let's divide numerator and denominator by x^2:
limx(x2+xy)\lim_{x \to \infty}\left(x^{2} + x y\right) =
limx(1+yx1x2)\lim_{x \to \infty}\left(\frac{1 + \frac{y}{x}}{\frac{1}{x^{2}}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
limx(1+yx1x2)=limu0+(uy+1u2)\lim_{x \to \infty}\left(\frac{1 + \frac{y}{x}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u y + 1}{u^{2}}\right)
=
0y+10=\frac{0 y + 1}{0} = \infty

The final answer:
limx(x2+xy)=\lim_{x \to \infty}\left(x^{2} + x y\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
limx(x2+xy)=\lim_{x \to \infty}\left(x^{2} + x y\right) = \infty
limx0(x2+xy)=0\lim_{x \to 0^-}\left(x^{2} + x y\right) = 0
More at x→0 from the left
limx0+(x2+xy)=0\lim_{x \to 0^+}\left(x^{2} + x y\right) = 0
More at x→0 from the right
limx1(x2+xy)=y+1\lim_{x \to 1^-}\left(x^{2} + x y\right) = y + 1
More at x→1 from the left
limx1+(x2+xy)=y+1\lim_{x \to 1^+}\left(x^{2} + x y\right) = y + 1
More at x→1 from the right
limx(x2+xy)=\lim_{x \to -\infty}\left(x^{2} + x y\right) = \infty
More at x→-oo