I have to compute the following limit $$\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}\sqrt{x^2+1}-\sqrt{x^2+x}$$ I have rewritten the function $f(x)$ as $$f(x)=\frac{x^2+1-x^2-x}{\sqrt{x^2+1}+\sqrt{x^2+x}}=-\frac{1}{2}$$
Now since $\sqrt{x^2+1}\sim x$ and $\sqrt{x^2+x}\sim x$, as $x\to\infty$, if I consider these asymptotic equalities I could write: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}x-x=0$$ but in such a way the result is wrong...What is wrong in the second idea of resolution?
Your first method is fine, but if you want to use asymptotic expansion, you have to consider higher terms. Recall that $(1+t)^a\sim1+at$ as $t\to 0$: $$\sqrt{x^2+1}-\sqrt{x^2+x}=x\left(\sqrt{1+\frac1{x^2}}-\sqrt{1+\frac1{x}}\right)\sim x\left(1+\frac1{2x^2}-1-\frac1{2x}\right)\to-\frac12 $$