$\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}$ by L'Hospital rule Consider evaluation of the limit $\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}$. By a direct way, we have
$$
\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}=\lim_{x\rightarrow\infty} \frac{1}{\sqrt{x+1}+\sqrt{x}}=0.
$$
On the other hand, if $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{x}$, then $\lim_{x\rightarrow\infty} f(x)=\lim_{x\rightarrow\infty} g(x)=\infty$, so the above limit is in indeterminate form $\infty-\infty$. If we want to apply L'Hospital's rule, we try to convert it into $\infty\cdot 0$ form by
$$f(x)-g(x)=f(x)g(x)\Big{(} \frac{1}{g(x)}-\frac{1}{f(x)}\Big{)}=\sqrt{x+1}\sqrt{x} \Big{(} \frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\Big{)}.$$
But, after this, I was unable to compute the limit of this by L'Hospital's rule. Any hint for it?

No problem!
$$\lim_{x\rightarrow+\infty}(\sqrt{x+1}-\sqrt{x})=\lim_{x\rightarrow+\infty}\frac{x}{\sqrt{x^3+x^2}+\sqrt{x^3}}=\lim_{x\rightarrow+\infty}\frac{1}{\frac{3x^2+2x}{2\sqrt{x^3+x^2}}+\frac{3}{2}\sqrt{x}}=0.$$
