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 $2$.

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 $.
