Evaluating $\lim\limits_{x\to \infty} \sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}$

Question

I cannot solve the following limit with radicals:

$$\lim_{x\to \infty} \sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}. $$

Answer 1

Multiply by $$\dfrac{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$ This gives us: $$\lim_{x\to \infty} \Big(\sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}\Big)\cdot \dfrac{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}} $$

$$ = \lim_{x\to \infty} \dfrac{(x^2 - 3x + 5) - (x^2 + 2x + 1)}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$

$$ = \lim_{x\to \infty} \dfrac{-5x + 4}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$ Now, try to evaluate.

You should obtain a limit of $$\dfrac{-5}{2}$$

Answer 2

Setting $x=\frac1h$

$$\lim_{x\to\infty}(\sqrt{x^2-3x+5}-\sqrt{x^2+2x+1})$$

$$=\lim_{h\to0^+}\frac{\sqrt{1-3h+5h^2}-\sqrt{1+2h+h^2}}h$$

$$=\lim_{h\to0^+}\frac{(1-3h+5h^2)-(1+2h+h^2)}{h(\sqrt{1-3h+5h^2}+\sqrt{1+2h+h^2})}$$

$$=\lim_{h\to0^+}\frac{4h^2-5h}h\cdot\frac1{\lim_{h\to0}\sqrt{1-3h+5h^2}+\sqrt{1+2h+h^2}}$$

Now as $h\to0^+,h\ne0$ the limit reduces to

$$\lim_{h\to0^+}\frac{4h-5}1\cdot\frac1{\lim_{h\to0}\sqrt{1-3h+5h^2}+\sqrt{1+2h+h^2}}$$

$$=-5\cdot\frac1{1+1}$$