# How can one find this limit?

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.

• 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. – Isomorphic Dec 11 '13 at 12:56
• You're missing a parantheses. HALP – mdenton8 Dec 11 '13 at 12:58

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