Difference of a root and a linear term

Text

Compute $\displaystyle\lim_{x\to +\infty}\left(\sqrt{x^{2}+3x}-x\right)$.

Solution

This is the indeterminate form $\infty-\infty$. Rationalize by multiplying and dividing by the conjugate: $\sqrt{x^{2}+3x}-x=\frac{\left(\sqrt{x^{2}+3x}-x\right)\left(\sqrt{x^{2}+3x}+x\right)}{\sqrt{x^{2}+3x}+x}=\frac{3x}{\sqrt{x^{2}+3x}+x}.$ Divide top and bottom by $x>0$ (so $x=\sqrt{x^{2}}$): $\frac{3x}{\sqrt{x^{2}+3x}+x}=\frac{3}{\sqrt{1+\tfrac{3}{x}}+1}\xrightarrow[x\to+\infty]{}\frac{3}{1+1}=\frac32.$ $\boxed{\,L=\tfrac32\,}$