Calculus problem with negative infinity $$\lim_{x\to-\infty} x(\sqrt{x²+1}-x)$$
Can someone simplify this and explain the steps. I'm having problems getting rid of sqrt.
 A: No need to get rid of the square root.  As $x\to -\infty$, both $\sqrt{x^2+1}$ and $-x$ go to $+\infty$, and thus the product goes to $-\infty$.
A: For $x \gt 0$:  $x \lt \sqrt{x^2+1} \lt  x+\tfrac{1}{2x}$ as you can check by squaring both sides, with the right-hand expression being very close for large $x$.  
So for $x \lt 0$:  $-x \lt \sqrt{x^2+1} \lt  -x-\tfrac{1}{2x}$
So for $x \lt 0$: $-2x^2  \gt  x(\sqrt{x^2+1}-x) \gt -2x^2 - \frac12$ 
and so the limit as $x \to -\infty$ is $-\infty$, and for large negative $x$, we have $-2x^2 - \frac12$ as an excellent lower bound
A: $\displaystyle \lim_{x \to -\infty}\sqrt{x^2+1}=\lim_{x \to -\infty}\sqrt{x^2(1+1/x^2)}=\lim_{x \to -\infty}|x|\sqrt{1+1/x^2}=\lim_{x \to -\infty}|x|\sqrt{1}=\lim_{x \to -\infty}|x|$
$$\lim_{x \to -\infty}x(\sqrt{x^2+1}-x)=\lim_{x \to -\infty}x(\sqrt{x^2(1+1/x^2)}-x)=\ \lim_{x \to -\infty}x(|x|-x)=\lim_{x \to -\infty}x((-x)-x)=\lim_{x \to -\infty}-2x^2=-\infty$$
$\displaystyle \lim_{x \to -\infty}|x|=\lim_{x \to -\infty}(-x)$ because by definition of $|x|$ we know that $|x|=-x$ when $x<0$ and in this case because $x$ is approaching $-\infty$ we know $x<0$.
as a general rule, if you have an algebraic expression and $x \to \infty$ only the highest degree term will matter, because you can factor the highest degree term and you'll have two factors, one is the highest degree term and the other goes to $1$ Like the following case:
$\displaystyle \lim_{x \to \infty}a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=\lim_{x \to \infty}a_nx^n(1+\frac{a_{n-1}}{a_nx}+\cdots+\frac{a_1}{a_nx^{n-1}}+\frac{a_0}{a_nx^n})=\lim_{x \to \infty}a_nx^n.\lim_{x \to \infty}(1+\frac{a_{n-1}}{a_nx}+\cdots+\frac{a_1}{a_nx^{n-1}}+\frac{a_0}{a_nx^n})=\lim_{x \to \infty}a_nx^n.1=\lim_{x \to \infty}a_nx^n$
A: At infinity sqrt( x^2+1) tends to x. So we have for the limitation process something like Lim x (x - x) as x tends to zero. I think it is an undefined condition and says nothing ...
