Źródło TeX:

\underset{x\rightarrow +\infty}{\lim }\,\dfrac{\sqrt{x^2+1}}{x}=\left[\dfrac{\infty}{\infty}\right]\overset{\text{H}}{=}\underset{x\rightarrow +\infty}{\lim }\,\dfrac{\frac{1}{2\sqrt{x^2+1}}\cdot 2x}{1}=\underset{x\rightarrow +\infty}{\lim }\,\dfrac{x}{\sqrt{x^2+1}}\overset{\text{H}}{=}\underset{x\rightarrow+ \infty}{\lim }\,\dfrac{\sqrt{x^2+1}}{x}