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If I am given the following graph, how do I compute the limit of x->2 of f(f(x))?

I tried to "compose" the limits, but lim x->2 f(x) is 1, but then f is not continuous at 1, so the limit DNE? Is that right?

enter image description here

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That is correct. –  par Dec 16 '13 at 3:02

2 Answers 2

up vote 10 down vote accepted

No, that is not correct. From the graph, no matter how $x$ approaches $2$, $f(x)$ will approach $1$ from below. Therefore, to find the limit of $f(f(x))$, you want the one-sided limit of $f$ at $1$, which does exist and can be seen on the graph to be $-1$.

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Great, thanks! I think that then I can also use this "from above" notion to solve the same limit as x approaches -1 (and get the answer being 2). But what about as x approaches 4? Then the above/below/left/right limits all go to 1, and it's not clear which one-sided limit to use –  MikeS Dec 16 '13 at 4:25
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@MikeS: Draw a graph of $f(f(x))$ for $x$ between $3.5$ and $4.5$, and you will be enlightened. –  Henning Makholm Dec 16 '13 at 4:28
    
@MikeS: Wait, how do you get 2 for $\lim_{x\to -1}f(f(x))$? –  Henning Makholm Dec 16 '13 at 4:32
    
Since as x goes to -1, f(x) gores to 1 from above. so the limit is the limit of f(x) for x going to 1 from above? –  MikeS Dec 16 '13 at 4:35
    
@MikeS: Yes. But how does that become $2$? –  Henning Makholm Dec 16 '13 at 4:36

As $x \to 2$ from the left, $f(x) \to 1$ from below; therefore, $$\lim_{x \to 2^-} f(f(x)) = \lim_{x \to 1^-} f(x) = -1$$

But as $x \to 2^+$, we have the exact same behavior: $f(x)$ approaches $1$ from below, so the limit is again $-1$. It's possible to make such limit computations since $f$ is continuous on an interval to the left of $1$.

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