# Another limit task, I've multiplied by the conjugate, now what?

The task: $\lim_{x\to\infty} \sqrt{x^2+1} -x$

I've multiplied with the conjugate expression ($\sqrt{x^2+1} +x$), then I get this

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

Is this correct so far? And what would be the next step?

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No, $(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)$ isn't $x^2+2$. – Robin Chapman Nov 20 '10 at 18:59
Fixed. Thanks!. – Algific Nov 20 '10 at 19:11

Remember that $a^{2}-b^{2}=(a-b)(a+b)$, put $a=\sqrt{x^{2}+1}$ and $b=x$ get:

$$(\sqrt{x^{2}+1}+x)(\sqrt{x^{2}+1}-x)=(\sqrt{x^{2}+1})^{2}-(x)^{2} = x^{2}+1-x^{2}= 1.$$

The end...

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The multiplication of conjugates is not correct: $(\sqrt{x^2+1}+x)(\sqrt{x^2+1}-x)=(x^2+1)-x^2=1$

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if you put braces around the thing after \sqrt then it all goes under the radical sign – Ross Millikan Nov 20 '10 at 19:03
@Ross: Thanks for the edit. I use softwares for latexing my stuff that don't need typing up latex formulae, so I tend to forget them from time to time. – Timothy Wagner Nov 20 '10 at 19:08

We have $$\sqrt{1 + \epsilon} = 1 + \frac{\epsilon}{2} + O(\epsilon^2)$$ using the Taylor expansion. So $$\sqrt{x^2+1} = x\sqrt{1+x^{-2}} = x\left(1 + \frac{1}{2x^2} + O(x^{-4})\right) = x + \frac{1}{2x} + O(x^{-3}).$$ Hence $$\sqrt{x^2+1} - x = \frac{1}{2x} + O(x^{-3}) \longrightarrow 0.$$

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HINT: What you have to do is this : $$\sqrt{x^{2}+1} -x \times \frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1} + x} = \frac{1}{\sqrt{x^{2}+1} +x}$$ then proceed to evaluate your limit.

Now observe that $\lim_{x \to \infty} \frac{1}{x} = 0$.

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$$\underset{x\rightarrow \infty }{\lim }\sqrt{x^{2}+1}-x=\underset{% x\rightarrow \infty }{\lim }\frac{\left( \sqrt{x^{2}+1}-x\right) \left( \sqrt{x^{2}+1}+x\right) }{\sqrt{x^{2}+1}+x}=\underset{x\rightarrow \infty }{% \lim }\frac{1}{\sqrt{x^{2}+1}+x}$$

$$\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}=\infty$$

$$\underset{x\rightarrow \infty }{\lim }\frac{1}{\sqrt{x^{2}+1}+x}=\dfrac{\underset{x\rightarrow \infty }{\lim }1}{\underset{x\rightarrow \infty }{\lim }{\sqrt{x^{2}+1}+x}}=\dfrac{1}{\infty}=0$$

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Thats what basically i have done. – anonymous Nov 20 '10 at 19:08
The limit is zero, but how do you actually show it? – Algific Nov 20 '10 at 19:10
@Chandru1: That's right! When I started writting my answer there was only Timothy Wagner's one. – Américo Tavares Nov 20 '10 at 19:10
@Algific: The limit of the denominator is $\infty$. – Américo Tavares Nov 20 '10 at 19:11
@Algific: I added the computation using the quocient rule of the limit. – Américo Tavares Nov 20 '10 at 19:28

HINT $\rm\ \ \ \$ Put $\rm\ f = x,\ g = 1\$ in the following simple

LEMMA $\rm\displaystyle\ \ \ \sqrt{f^2+g}\ -\ f\ \to\ 0 \ \ \ if\ \ \ \frac{f}g \to\ \infty\ \ \ and\ \ \frac{1}g\ \to\ \infty\ \ or\ \ r\in \mathbb R$

Proof $\rm\displaystyle\quad\quad\ \ \sqrt{f^2+g}\ -\ f\ \ =\ \frac{g}{\sqrt{f^2+g}\ +\ f}\ =\ \frac{1}{\sqrt{(\frac{f}g)^2+\frac{1}g}\ +\ \frac{f}g}$

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