Compute $\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)$ I note that $\sqrt{x^2-6x+9}=|x-3|$. Splitting upp the limit into cases gives


*

*$x\geq 3:$


$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(|x-3|+x-1)=2\lim_{x\rightarrow -\infty}(x-2)=-\infty.$$


*

*$x< 3:$


$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
I get two different values of the limit, but plotting the function clearly shows that the answer should be $2$.
 A: Put $t=-x$, then: 
$$\lim_{t\rightarrow \infty}\left(\sqrt{t^2+6t+9}-t-1\right) = \lim_{t\rightarrow \infty}\left(|t+3|-t-1\right) =\lim_{t\rightarrow \infty}\left(t+3-t-1\right) = 2
 $$
A: You're approaching $-\infty$, so take, 
$$|x-3| = 3-x$$
(I don't get where you're going with $x \ge 3$)
$$\lim_{x \to -\infty} \sqrt{(x-3)^2} + x - 1$$
$$= \lim_{x \to -\infty} |x-3| + x - 1$$
$$= \lim_{x \to -\infty} 3 - x + x - 1$$
$$= \lim_{x \to -\infty} 2$$
$$= \boxed 2$$
A: Now, we have $\lim_{x\rightarrow -\infty}$ which implies x is approaching the negative infinity.
if $x\geq 3$, x is still on the way approaching the infinity from somewhere and x needs to pass the domain of $x\geq 3$ Then it goes to the domain of $x< 3$ therefore
$$\lim_{x\rightarrow -\infty}\left(\sqrt{x^2-6x+9}+x-1\right)=\lim_{x\rightarrow -\infty}(-x+3+x-1)=\lim_{x\rightarrow -\infty}2=2.$$
A: Here is a different approach, setting $x=-t$, the limit becomes:
$$\lim_{t \to \infty} \sqrt{t^2+6t+9} - t - 1$$
Since $t^2+6t+9=(t+3)^2$, we can say for positive t-values that $\sqrt{(t+3)^2}=t+3$, your limit expression becomes $t+3-t-1=2$.
A: Like you said there is 2 cases: $$\begin{cases}x\ge 3\\x<3\end{cases}$$
Now you tell me: is $-∞\ge 3$ or $-∞<3$?
A: Since $\lim_{x \rightarrow - \infty}$ it suffices to consider 'small' negative $x$.
$|x-3| = -x + 3$.
Example: $x= -7$: 
$|x-3| = |-7-3| = -x +3.$
Hence: 
$\lim_{x \rightarrow -\infty} (|x-3| + x-1) = $
$\lim_{x \rightarrow -\infty}( -x +3 +x -1 )= 2 $.
