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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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    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.2012-11-14

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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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    @knot: My apologies for casting aspersions on your algebra, which was fine: I misread one of the signs.2012-11-14