0
$\begingroup$

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 ?

  • 0
    The question asked for $\lim_{x\to\infty}$. Why are you worried about what happens when $x\to-\infty$?2012-11-14
  • 0
    I think that $\infty$ can be $+\infty$ or $-\infty$2012-11-14
  • 0
    @knot: But what matters is what the person who asked the question thinks.2012-11-14
  • 0
    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

1 Answers 1