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.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ 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... $\endgroup$ – Avitus Dec 11 '13 at 12:55
-
$\begingroup$ @Avitus You can't assume such a thing. $\endgroup$ – Isomorphic Dec 11 '13 at 12:56
-
$\begingroup$ You're missing a parantheses. HALP $\endgroup$ – mdenton8 Dec 11 '13 at 12:58
Add a comment
|
$\begingroup$
$\endgroup$
9
$$\lim_{x\to 2} (f^2(x)-6f(x)+9)=0\iff\lim_{x\to 2} (f(x)-3)^2=0$$
-
$\begingroup$ I've reached that point. I'm not sure why this leads to the conclusion that $\lim_{x\rightarrow 2}f(x)=3$ $\endgroup$ – Matheo Dec 11 '13 at 12:56
-
1$\begingroup$ 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? $\endgroup$ – Brian Dec 11 '13 at 12:58
-
1$\begingroup$ @Matheo Because $\lim_{x\to 2} (f(x)-3)^2=(\lim_{x\to 2} (f(x)-3))^2=0<=>\lim_{x\to 2} (f(x)-3)=0=>\lim_{x\to 2} f(x)-\lim_{x\to 2} 3=0<=>\lim_{x\to 2} f(x)=\lim_{x\to 2} 3=3$ $\endgroup$ – Haha Dec 11 '13 at 13:03
-
$\begingroup$ @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. $\endgroup$ – mdenton8 Dec 11 '13 at 13:04
-
1$\begingroup$ @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. $\endgroup$ – Haha Dec 11 '13 at 13:10
