Limit problem

Observe that when $x$ is close to $6$ but smaller ( $5.999$ ), we can see that $f(x)$ is close to $2$ , but bigger.

And when $x$ is close to $6$ but bigger ( $6.001$ ), we can see that $f(x)$ is close to $2$ , but bigger.

So we can see that whenever $x$ approaches $6$ , in both sides $f(x)$ approaches $y$ -value of $2$ from above.

So $\displaystyle \lim_{x\to6}f(f(x)) = \lim_{x\to2^+}f(x) = \boxed{5}$

Approach 1: As $x \rightarrow 6$ , we have $f(x) \rightarrow 2$ . However, the way that $f(x)$ approaches 2 is only on the positive side. Thus, it is more accurate to say that $f(x) \rightarrow 2^+$ . Hence,

$\lim_{x \rightarrow 6} f(f(x) ) = \lim_{y \rightarrow 2^+} f(y) = 5$

Approach 2: Let's create a table for $x, f(x), f(f(x))$ (approximate values).

From the table, we see that the limit is 5.

For a limit to exist we need the limits from either side to exist, and to be equal.

Since $\frac{\mathrm{d}f}{\mathrm{d}x} < 0$ as $x \to 6^{-}$ , the left hand limit is: $\begin{aligned} \lim_{x \to 6^{-}} f(f(x)) &= \lim_{x \to 2^{+}} f(x) \\ &= 5 \end{aligned}$

Since $\frac{\mathrm{d}f}{\mathrm{d}x} > 0$ as $x \to 6^{+}$ , the right hand limit is: $\begin{aligned} \lim_{x \to 6^{+}} f(f(x)) &= \lim_{x \to 2^{+}} f(x) \\ &= 5 \end{aligned}$

Therefore, $\lim_{x \to 6} f(f(x)) = \boxed{5}$ .

For any who wonder why this is an accurate limit: Remember that limits are defined by small open intervals, later called open sets. The limit does $\color{#D61F06}\text{not}\color{#333333}$ depend on $f(x_0)$ :

$\begin{aligned} "\lim_{x\rightarrow x_0}f(x) &= y_0" & \Leftrightarrow && "\text{For any (small) }\varepsilon>0:\quad|f(x) - y_0|<\varepsilon \quad\text{ as long as } \quad\color{#D61F06} 0< \color{#333333}|x-x_0|<\delta" \end{aligned}$

Now to the task: We have $x_0=6$ , guess $y_0=5$ and create any small interval $I_y:=(5-\varepsilon,\:5+\varepsilon)$ around $y_0=5$ . Let's construct a small interval $I_x:=(6-\delta,\:6+\delta)$ using the graphic by the following steps:

1. Take any (small) $\varepsilon>0$ and find a function value $5-\varepsilon < y_1 < 5$ .
2. Find a corresponding $x$ -value slightly bigger than 2 such that $f(x_1)=y_1$
3. Make $\delta>0$ small enough, so that for any $\begin{aligned} 0<|x-6|<\delta:&&&|f(x)-2| < |x_1-2| \end{aligned}$

By construction, the following statement holds: $\begin{aligned} "0<|x-6|<\delta&&\Rightarrow&&|f(f(x))-5|<\varepsilon"&&\Rightarrow&&\lim_{x\rightarrow 6}f(f(x))=\fbox{5} \end{aligned}$

Rem.: You cannot argue with differentiation or continuity here, because $f$ is not continuous in $x=2$ . However, for a limit to exist we only need to construct these small open intervals, regardless of what properties the function might have.