Convergence Proof: $\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}$ [duplicate]

Hint: Multiply by $\frac{\sqrt{4x+x^2}+\sqrt{x^2+x}}{\sqrt{4x+x^2}+\sqrt{x^2+x}}$.

You get nowhere by reasoning that way. The canonical way to work on this is "rationalizing": you multiply and divide by $\sqrt{4x+x^2}+\sqrt{x^2+x}$. That way you get $$ \sqrt{4x+x^2}-\sqrt{x^2+x}=\frac{4x+x^2-x^2-x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac{3x}{\sqrt{4x+x^2}+\sqrt{x^2+x}}=\frac3{\sqrt{4/x+1}+\sqrt{1+1/x}}\to\frac32 $$

$\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x} = \lim_{x\rightarrow\infty} \frac{(\sqrt{4x+x^2}- \sqrt{x^2+x})\cdot(\sqrt{4x+x^2}+ \sqrt{x^2+x})}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{4x+x^2-x^2-x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{3x}{ \sqrt{4x+x^2}+ \sqrt{x^2+x}} = \lim_{x\rightarrow\infty} \frac{3}{ \sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{ \lim_{x\rightarrow\infty}\sqrt{\frac{4}{x}+1}+ \sqrt{1+\frac{1}{x}}} = \frac{3}{\sqrt{1}+\sqrt{1}} =\frac{3}{2}$