Inspired by Khan

Multiplying with conjugate $\sqrt{x^2+4x+1}+x$ on both parts of fraction gives $\frac{(x^2+4x+1)-x^2}{\sqrt{x^2+4x+1}+x} = \frac{4x+1}{\sqrt{x^2+4x+1}+x}$ then divide with $x$ on both parts gives $\frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1}$ As $lim_{x\rightarrow\infty}\frac{1}{x} = 0$ , the limit of the above function is then $\frac{4+0}{\sqrt{1+0+0}+1} = \frac{4}{2} = 2$

I factored x out to yield: $x\cdot \left(\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}-1\right)$ Then rearranged to divide by 1/x and applied L'Hôpital's rule. It got ugly but yielded 2.