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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 $2$.

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  • $\begingroup$ Are you sure that you can reach $-\infty$ with values of $x$ over $3$ ? $\endgroup$ – Yves Daoust Sep 22 '17 at 13:41
  • $\begingroup$ I know that it is incorrect, but I can't se where the error is. $\endgroup$ – Parseval Sep 22 '17 at 13:42
  • $\begingroup$ $$x\to-\infty \implies. x<3$$ $\endgroup$ – lab bhattacharjee Sep 22 '17 at 13:43
  • $\begingroup$ I just told you but I am not sure you took care... $\endgroup$ – Yves Daoust Sep 22 '17 at 13:43
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    $\begingroup$ If $x$ tends to negative infinity, do we need to consider $x\geq 3$? $\endgroup$ – Jihoon Kang Sep 22 '17 at 13:43
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Like you said there is 2 cases: $$\begin{cases}x\ge 3\\x<3\end{cases}$$ Now you tell me: is $-∞\ge 3$ or $-∞<3$?

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Put $t=-x$, then: $$\lim_{t\rightarrow \infty}\left(\sqrt{t^2+6t+9}-t-1\right) = \lim_{t\rightarrow \infty}\left(|t+3|-t-1\right) =\lim_{t\rightarrow \infty}\left(t+3-t-1\right) = 2 $$

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  • $\begingroup$ Why should I do that? Why doesn't my way work? $\endgroup$ – Parseval Sep 22 '17 at 13:49
  • $\begingroup$ @John. How does your method differ from mine? $\endgroup$ – imranfat Sep 22 '17 at 22:00
  • $\begingroup$ @imranfat: It does not and I could ask you the same question. $\endgroup$ – Maria Mazur Sep 22 '17 at 22:02
  • $\begingroup$ @johnnobody. In that case, you should receive an upvote :) $\endgroup$ – imranfat Sep 22 '17 at 22:04
  • $\begingroup$ Yes, I suppose you are right. $\endgroup$ – Maria Mazur Sep 22 '17 at 22:05
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You're approaching $-\infty$, so take,

$$|x-3| = 3-x$$

(I don't get where you're going with $x \ge 3$)

$$\lim_{x \to -\infty} \sqrt{(x-3)^2} + x - 1$$

$$= \lim_{x \to -\infty} |x-3| + x - 1$$

$$= \lim_{x \to -\infty} 3 - x + x - 1$$

$$= \lim_{x \to -\infty} 2$$ $$= \boxed 2$$

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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

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

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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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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 $.

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