# Limit of a function at infinity

I have a question that finding the limit : $\text{lim}_{x\rightarrow \infty}x(\sqrt{x^2+1}-x)$.

My strategy is follows :

$\text{lim}_{x\rightarrow \infty}x(\sqrt{x^2+1}-x)=\text{lim}_{x\rightarrow \infty}\dfrac{x}{\sqrt{x^2+1}+x}$

From this if I divide both the denominator and the numerator by $x$, then it wil depend whether $x\rightarrow +\infty$ or $x\rightarrow -\infty$ to conclude and two case wil give two answer $1$ and $-1$.

So, am I wrong any where ? How can I solve it ?

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The question asked for $\lim_{x\to\infty}$. Why are you worried about what happens when $x\to-\infty$? –  Gerry Myerson Nov 14 '12 at 3:39
I think that $\infty$ can be $+\infty$ or $-\infty$ –  knot Nov 14 '12 at 3:40
@knot: But what matters is what the person who asked the question thinks. –  André Nicolas Nov 14 '12 at 3:48
When someone says $\lim_{x\rightarrow \infty} f(x) = L$, they mean the following: $\forall \epsilon > 0, \exists \delta >0$ such that $x \in (\delta,\infty) \Rightarrow |f(x)-L|< \epsilon$. Thus $+\infty$ and $-\infty$ are two different things. –  Gautam Shenoy Nov 14 '12 at 4:46

Corrected: Presumably you got

$$x\left(\sqrt{x^2+1}-x\right)=\frac{x}{\sqrt{x^2+1}+x}$$

by some version of the trick of multiplying by $1$ in a carefully chosen disguise. To continue, do it again, but this time with the disguise $1=\dfrac{1/x}{1/x}$, using the fact that $\sqrt{x^2+1}=\sqrt{1+\frac1{x^2}}$ for positive $x$:

\begin{align*} \frac{x}{\sqrt{x^2+1}+x}&=\frac{x}{\sqrt{x^2+1}+x}\cdot\frac{1/x}{1/x}\\\\ &=\frac1{\sqrt{1+\frac1{x^2}}+1} \end{align*}

for $x>0$. (Since we’re going to take the limit as $x\to\infty$, we care only about $x>0$.) Now go ahead and take the limit as $x\to\infty$.

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Sorry, but how did you get $\sqrt{x^2+1}=x\sqrt{1+\dfrac{1}{x^2}}$ ? If we take $x$ out of $\sqrt{x^2+1}$, I think we have to care about the sign of $x$. –  knot Nov 14 '12 at 3:51
@knot: Yes, I should really have written $|x|$, but we’re interested in the limit as $x\to\infty$, so we’re interested only in positive $x$. I’ll revise it slightly to make that clear. –  Brian M. Scott Nov 14 '12 at 3:54
@Gerry: Ouch. Indeed. –  Brian M. Scott Nov 14 '12 at 4:51
@knot: My apologies for casting aspersions on your algebra, which was fine: I misread one of the signs. –  Brian M. Scott Nov 14 '12 at 4:54

$\displaystyle \lim_{x \rightarrow \infty} \left[ x \left( \sqrt{x^2 + 1} - x \right) \right] = \lim_{x \rightarrow \infty} \left[ x \left( x\sqrt{1 + \frac 1{x^2}} - x \right) \right] = \lim_{x \rightarrow \infty} \left[ x^2 \left( \sqrt{1 + \frac 1{x^2}} - 1 \right) \right] = \lim_{x \rightarrow \infty} \left[ x^2 \left( 1 + \frac 1{2x^2} - 1 \right) \right] = \frac 12$

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It'd be great if you explain how you pass from the LHS to the middle term, justifying the step of course. –  DonAntonio Nov 14 '12 at 13:21
It's a power series expansion of square root function after taking $x^2$ out of the root as $x$, with assumption $x>0$ due to the $x \rightarrow \infty$ of course. Anyway, edited my answer with a bit more details. –  Kaster Nov 14 '12 at 19:11