Limits problem $\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}$ The question is to find:
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}$$
Working number 1:
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}=\lim_{x\rightarrow \infty }(\sqrt{x^2+x}-x)\times\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x} $$
$$\lim_{x\rightarrow \infty }\frac{(\sqrt{x^2+x})^2-x^2}{\sqrt{x^2+x}+x}=\lim_{x\rightarrow \infty }\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\lim_{x\rightarrow \infty }\frac{x}{\sqrt{x^2+x}+x}$$
$$\lim_{x\rightarrow \infty }\frac{\frac{x}{\sqrt{x}^2}}{\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}+\frac{x}{\sqrt{x}^2}}=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}$$
Working number 2: 
$$\lim_{x\rightarrow \infty }{\sqrt{x^2+x}-x}=\lim_{x\rightarrow \infty }\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}-\frac{x}{\sqrt{x}^2}=\sqrt{1+0}-1=0$$
The answer was $\frac{1}{2}$, but the question is:
What is wrong with working number 2? 

Your first solution is fine but the second solution is a bit problematic. 
$$\sqrt{\frac{x^2}{{x^2}}+\frac{x}{{x}^2}}-\frac{x}{\sqrt{x}^2} = \frac{\sqrt{x^2+x}}{x} - 1 \neq {\sqrt{x^2+x}-x}$$
