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

I cannot solve the following limit with radicals:

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

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## closed as off-topic by Dan Rust, Davide Giraudo, Carl Mummert, Eric Naslund, Daniel Robert-NicoudNov 25 '13 at 16:42

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– Eric Naslund Nov 25 '13 at 16:05

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 $$\dfrac{-5}{2}$$

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I'm getting -5x+4 in nominator and radicals in denominator. Still inf/inf – RCola Nov 25 '13 at 14:55
Yes, RCola. That's correct. You can divide numerator and denominator by $x$. In the denominator, dividing by $x$ is the same thing as dividing by $\sqrt{x^2}$. This gives you $$\frac{-5 + \frac 4x}{\sqrt{1 - \frac 3x + \frac 5{x^2}} + \sqrt{1 + \frac 2x + \frac 1{x^2}}}$$ and the limit is then $$\dfrac {-5 + 0}{\sqrt{1 + 0+0} + \sqrt{1 + 0 + 0}} = \frac{-5}{2}$$ – amWhy Nov 25 '13 at 14:59
@amWhy: Just got home! Needs another UV +_1 – Amzoti Nov 26 '13 at 3:56

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

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

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

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