<s>[INST] 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 $2$.
 [/INST] 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 $t+3-t-1=2$.
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