Useful theorems/properties for evaluating $\lim \limits_{x \to +\infty}\sqrt{x}(\sqrt{x+1}-\sqrt{x-1})$ I have been trying to solve the following question from my Mathematical Analysis course in university:
$$\lim \limits_{x \to +\infty}\sqrt{x}(\sqrt{x+1}-\sqrt{x-1})$$
I am aware that the answer is 1, but I am not entirely sure why.
Substituting positive infinity into the equation, if I am not terribly mistaken, gives $\infty-\infty$, which is indeterminate. Anyhow, I am not aware of any properties or theorems that could get rid of this issue, and Landau symbols don't really seem to help. I have tried as well to rewrite the problem as $$\lim \limits_{x \to +\infty}\frac{\sqrt{x+1}-\sqrt{x-1}}{\frac{1}{\sqrt{x}}}$$
to use L'Hopital's theorem, which does not get rid of the indeterminate form. Help would be appreciated. Thanks in advance!
 A: In fact Landau symbols do work very well.
$\sqrt{1\pm u}=1\pm\frac 12 u+o(u)\ $ for $\ u\to 0$
So factor out $x$ to get:
$\begin{align}\require{cancel}x\left(\sqrt{1+\frac 1x}-\sqrt{1-\frac 1x}\right)&=x\Big((\cancel{1}+\frac 1{2x}+o(\frac 1x))-(\cancel{1}-\frac 1{2x}+o(\frac 1x))\Big)\\&=x\Big(\frac1x+o(\frac 1x)\Big)\\&=1+o(1)\\&\to 1\end{align}$
But you can as well use the conjugated quantity method as always when difference of square roots is used:
$\sqrt{x+1}-\sqrt{x-1}=\dfrac{(x+1)-(x-1)}{\sqrt{x+1}+\sqrt{x-1}}=\dfrac{2}{\sqrt{x+1}+\sqrt{x-1}}\sim \dfrac 2{2\sqrt{x}}\sim\dfrac 1{\sqrt{x}}$
And multiplied by $\sqrt{x}$ this indeed has limit $1$.
A: Note that if $f(x)=\sqrt{x(x-1)}$ this is equivalent to finding
$$\lim_{x\to\infty} f(x+1)-f(x)$$
Note that for $x>1$, $\sqrt{x^2-x}<\sqrt{x^2-x+.25}=x-.5$. Also it's not hard to prove that for any $\epsilon<0$, there exists $\delta$ such that $\forall x>\delta$, $\sqrt{x^2-x}>x-.5-\epsilon$. In particular this value of $\delta$ is $\delta=\frac{(.5+\epsilon)^2}{2\epsilon}$. So $\forall \epsilon>0$, we know that $\forall x>\delta$, $1-\epsilon<f(x+1)-f(x)<1+\epsilon$. So it follows that $\lim_{x\to\infty} f(x+1)-f(x)=1$.
