Question

Solve: $:\underset{x\to \infty }{lim}\left(\sqrt{x^2+x}-x\right)$

Answer 1

The initial form for the limit is indeterminate $\infty -\infty$

So, use the conjugate

$Wehave{lim}_{x\to \infty }\left(\sqrt{x^2+x}-x\right)$

Since,

$=\frac{\left(\sqrt{x^2+x}-x\right)}{1}.\frac{\left(\sqrt{x^2+x}+x\right)}{\left(\sqrt{x^2+x}+x\right)}=\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}-x}$

$\frac{x}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2}\sqrt{1+\frac{1}{x}}+x}\left(forx\ne \right)=\frac{x}{x\sqrt{1+\frac{1}{x}}+x}\left(forx>0\right)=\frac{x}{x\left(\sqrt{1+\frac{1}{x}}+1\right)}=\frac{1}{\sqrt{1+\frac{1}{x}}+1}\therefore {lim}_{x\to \infty }\frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}$

Hence, this is the answer.