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It's given that $$\lim_{x\rightarrow 2}(f(x)^2-6f(x))=-9$$.

How can one figure out $$\lim_{x\rightarrow 2}f(x)?$$

Excuse me, if this is too easy.

1 Answers1

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$$\lim_{x\to 2} (f^2(x)-6f(x)+9)=0\iff\lim_{x\to 2} (f(x)-3)^2=0$$

Did
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Haha
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  • I've reached that point. I'm not sure why this leads to the conclusion that $\lim_{x\rightarrow 2}f(x)=3$ –  Dec 11 '13 at 12:56
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    Remember that $x^{2}$ is a continuous function, so $0 = \lim_{x\to2}\left(f\left(x\right)-3\right)^{2} = \left(\lim_{x\to2}f\left(x\right)-3\right)^{2}$. What can you conclude from this? – Brian Dec 11 '13 at 12:58
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    @Matheo Because $\lim_{x\to 2} (f(x)-3)^2=(\lim_{x\to 2} (f(x)-3))^2=0<=>\lim_{x\to 2} (f(x)-3)=0=>\lim_{x\to 2} f(x)-\lim_{x\to 2} 3=0<=>\lim_{x\to 2} f(x)=\lim_{x\to 2} 3=3$ – Haha Dec 11 '13 at 13:03
  • @Brian Scholl. Isn't it more important that $\sqrt{x}$ is a continuous function on $[0, \infty)$ and therefore if $\lim\limits_{y \to 2} y^2 = z$ then $\lim\limits_{y \to 2} \sqrt{y^2} = \sqrt{z}$... Wondering for my own interest. – mdenton8 Dec 11 '13 at 13:04
  • Oh, I get it, so $\lim\limits_{y \to 2} y = \lim\limits_{y \to 2} y $ and $x^2$ is continuous, therefore $\lim\limits_{y \to 2} y^2 = (\lim\limits_{y \to 2} y)^2 $... Cool – mdenton8 Dec 11 '13 at 13:07
  • @mdenton8 In this case, we only need the continuity of $x^{2}$ because $x^{2} = 0 \iff x = 0$. So, when you see that $\left(\lim_{x\to2}f\left(x\right)-3\right)^{2} = 0$ you can then conclude that $\lim_{x\to2}\left(f\left(x\right)-3\right) = 0$ and reach the conclusion from there. – Brian Dec 11 '13 at 13:07
  • @mdenton8, it's more than that. It works the other way too, but only at one point. – dfeuer Dec 11 '13 at 13:08
  • Thanks @Mitsos. Is this solution acceptable when we don't know whether the limit of the function exists? –  Dec 11 '13 at 13:08
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    @Matheo, if you mean the $\lim f(x)$ there is no problem. All we need to know is that $\lim_{x\to 2} (f(x)-3)^2$ exists. – Haha Dec 11 '13 at 13:10