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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.

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Let $f$ be continuous at $x=2$. The fist limit gives you a polynomial of degree 2 in the variable $f(2)$. Solve it and... –  Avitus Dec 11 '13 at 12:55
    
@Avitus You can't assume such a thing. –  Iota Dec 11 '13 at 12:56
    
You're missing a parantheses. HALP –  mdenton8 Dec 11 '13 at 12:58

1 Answer 1

up vote 14 down vote accepted

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

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I've reached that point. I'm not sure why this leads to the conclusion that $\lim_{x\rightarrow 2}f(x)=3$ –  Matheo 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 Scholl Dec 11 '13 at 12:58
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@Matheo Because $\lim_{x\to 2} (f(x)-3)^2=(\lim_{x\to 2} (f(x)-3))^2=0<=>\lim_{x\to 2} (f(x)-3)=0=>\lim_{x\to 2} f(x)-\lim_{x\to 2} 3=0<=>\lim_{x\to 2} f(x)=\lim_{x\to 2} 3=3$ –  Mitsos 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
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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. –  Mitsos Dec 11 '13 at 13:10

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