# 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