limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$

Question

How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.

$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$

Answer 1

Assume WLOG $x\geq0$. Thus, $$\frac{\sqrt{x+2}-\sqrt{x+1}}{\sqrt{x+1}-\sqrt x}\cdot\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+1}+\sqrt x}$$$$=(\sqrt{x+1}+\sqrt x)(\sqrt{x+2}-\sqrt{x+1})\cdot\frac{\sqrt{x+2}+\sqrt{x+1}}{\sqrt{x+2}+\sqrt{x+1}}$$$$=\frac{\sqrt{x+1}+\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}=1-\frac{\sqrt{x+2}-\sqrt x}{\sqrt{x+2}+\sqrt{x+1}}$$

Answer 2

\begin{gather*} \lim _{x\rightarrow \infty }\frac{\sqrt{x+2} -\sqrt{x+1}}{\sqrt{x+1} -\sqrt{x}}\\ =\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1}\\ \end{gather*} (Taking $\sqrt{x}$ common from both numerator and denominator)

Now, from the binomial series expansion, $(1+ay)^{b} =1+aby$, when $y\to 0$

\begin{gather*} \sqrt{1+\frac{2}{x}} =1+\frac{1}{2} \times \frac{2}{x}\\ \sqrt{1+\frac{1}{x}} =1+\frac{1}{2} \times \frac{1}{x}\\ \lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1} =\lim _{x\rightarrow \infty }\frac{\left( 1+\frac{1}{2} \times \frac{2}{x}\right) -\left( 1+\frac{1}{2} \times \frac{1}{x}\right)}{\left( 1+\frac{1}{2} \times \frac{1}{x}\right) -1} =1 \end{gather*}

Answer 3

Hint:

$$\sqrt{x+k+1}-\sqrt{x+k}=\frac{1}{\sqrt{x+k+1}+\sqrt{x+k}}\sim\dfrac1{2\sqrt x}$$