Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$

Question

$$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$

Can someone help me to solve it?

result of online calculator: 3/5

Answer 1

Hint $$\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)\left(\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+2}+\sqrt{x-3}}\right)=\frac{\big(\sqrt{x+1}-\sqrt{x-2}\big)\big(\sqrt{x+2}+\sqrt{x-3}\big)}{5}$$

Answer 2

Hint:

$$\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}} = \frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\cdot\frac{\sqrt{x+1}+\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-3}}\cdot\frac{\sqrt{x+2}+\sqrt{x-3}}{\sqrt{x+1}+\sqrt{x-2}}$$

Answer 3

Rewrite $$A=\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{\sqrt{x}\sqrt{1+1/x}-\sqrt{x}\sqrt{1-2/x}}{\sqrt{x}\sqrt{1+2/x}-\sqrt{x}\sqrt{1-3/x}}=\frac{\sqrt{1+1/x}-\sqrt{1-2/x}}{\sqrt{1+2/x}-\sqrt{1-3/x}}$$ Replace $\frac 1x$ by $y$; so $$A=\frac{\sqrt{1+y}-\sqrt{1-2y}}{\sqrt{1+2y}-\sqrt{1-3y}}$$ Now, use the fact that, when $z$ is small compared to $1$,$\sqrt{1+z}\approx 1+\frac z2$. Replace $z$ by the appropriate value for each radical.

I am sure that you can take from here.