Compute the limit $\lim_{x\to\infty} (\sqrt{x+3}-3\sqrt{x+1}+2\sqrt{x})$

Question

$$\lim_{x\to\infty} (\sqrt{x+3}-3\sqrt{x+1}+2\sqrt{x})=\lim_{x\to\infty} (x\sqrt{\frac{1}{x}+\frac{3}{x^2}}-3x\sqrt{\frac{1}{x}+\frac{1}{x^2}}+2x\sqrt{\frac{1}{x}})$$
I can see that we are in $\infty - \infty$ case. I just need a hint.

Answer 1

Hint: $$ \sqrt{x+3}-\sqrt{x+1} = \frac{2}{\sqrt{x+1}+\sqrt{x+3}}\to 0, \qquad \sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x+1}+\sqrt{x}}\to 0.$$

Answer 2

Set $1/x=h^2 \implies h\to0,$ WLOG $h>0,\sqrt{h^2}=+h$

$$\lim_{x\to\infty} (\sqrt{x+3}-3\sqrt{x+1}+2\sqrt{x})$$

$$=\lim_{h\to0^+}\dfrac{\sqrt{1+3h^2}-3\sqrt{1+h^2}+2}h$$

$$=\lim_{h\to0^+}\dfrac{\sqrt{1+3h^2}-1}h-3\cdot\lim_{h\to0^+}\dfrac{\sqrt{1+h^2}-1}h$$

Now $\lim_{h\to0^+}\dfrac{\sqrt{1+nh^2}-1}h=\lim_{h\to0^+}\dfrac{nh}{\sqrt{1+nh^2}+1}=0$ for finite $n$