Jeżeli x→±∞limf(x)=a\displaystyle\lim_{x\to \pm\infty}f(x)=a oraz x→±∞limg(x)=b\displaystyle\lim_{x\to\pm\infty}g(x)=b, gdzie a,b,∈Ra,b,\in\mathbb{R}, to:
x→±∞lim(c⋅f(x))=c⋅x→±∞limf(x)=c⋅a, gdzie c∈R\displaystyle\lim_{x\to \pm\infty}(c\cdot f(x))=c\cdot \lim_{x\to \pm\infty} f(x)=c\cdot a,\quad \text{ gdzie } c\in\mathbb{R}
x→±∞lim(f(x)+g(x))=x→±∞limf(x)+x→±∞limg(x)=a+b\displaystyle\lim_{x\to \pm\infty}(f(x)+g(x))=\lim_{x\to \pm\infty}f(x) + \lim_{x\to \pm\infty}g(x)=a+b
x→±∞lim(f(x)−g(x))=x→±∞limf(x)−x→±∞limg(x)=a−b\displaystyle\lim_{x\to \pm\infty}(f(x)-g(x))=\lim_{x\to \pm\infty}f(x) -\lim_{x\to \pm\infty}g(x)=a-b
x→±∞lim(f(x)⋅g(x))=x→±∞limf(x)⋅x→±∞limg(x)=a⋅b\displaystyle\lim_{x\to \pm\infty}(f(x)\cdot g(x))=\lim_{x\to \pm\infty}f(x) \cdot\lim_{x\to \pm\infty}g(x)=a\cdot b
x→±∞limg(x)f(x)=x→±∞limg(x)x→±∞limf(x)=ba, gdzie b=0\displaystyle\lim_{x\to \pm\infty}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to \pm\infty}f(x)}{\displaystyle\lim_{x\to \pm\infty}g(x)} =\frac{a}{b},\quad \text{ gdzie } b\neq0