Question

the figure above shows the graph of the function f. which of the following statements are true?
i. $lim_{x \to 2^-} f(x) = f(2)$
ii. $lim_{x \to 6^-} f(x) = \lim_{x \to 6^+} f(x)$
iii. $lim_{x \to 6} f(x) = f(6)$
options:
a) ii only
b) iii only
c) i and ii only

Explanation:

Step1: Analyze Statement I

To check \(\lim_{x
ightarrow2^{-}}f(x)=f(2)\), we look at the left - hand limit as \(x\) approaches \(2\) and the value of \(f(2)\). From the graph, as \(x\) approaches \(2\) from the left (\(x
ightarrow2^{-}\)), we can see the trend of the function. The value of \(f(2)\) is \(- 2\) (from the graph, at \(x = 2\), the function has a closed dot at \(y=-2\)). The left - hand limit as \(x
ightarrow2^{-}\): we follow the graph from the left side of \(x = 2\) towards \(x = 2\). The graph is a line approaching \(y=-2\) as \(x
ightarrow2^{-}\). So \(\lim_{x
ightarrow2^{-}}f(x)=-2\) and \(f(2)=-2\), so statement I is true.

Step2: Analyze Statement II

We need to check if \(\lim_{x
ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{+}}f(x)\). For the left - hand limit as \(x
ightarrow6^{-}\), we look at the graph as \(x\) approaches \(6\) from the left. The graph is a line approaching \(y = - 2\) (since before \(x = 6\), the function is decreasing towards \(x = 6\)). For the right - hand limit as \(x
ightarrow6^{+}\), we look at the graph as \(x\) approaches \(6\) from the right. The graph is a line that starts from \(x = 6\) (open dot at \(x = 6\), \(y=-2\)) and goes up. So \(\lim_{x
ightarrow6^{-}}f(x)=-2\) and \(\lim_{x
ightarrow6^{+}}f(x)=-2\), so \(\lim_{x
ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{+}}f(x)\), statement II is true.

Step3: Analyze Statement III

We need to check if \(\lim_{x
ightarrow6}f(x)=f(6)\). First, find \(\lim_{x
ightarrow6}f(x)\). Since \(\lim_{x
ightarrow6^{-}}f(x)=\lim_{x
ightarrow6^{+}}f(x)=-2\), then \(\lim_{x
ightarrow6}f(x)=-2\). Now, look at \(f(6)\): from the graph, at \(x = 6\), there is an open dot (so the function is not defined at \(x = 6\) with the value of the open dot, but wait, actually, the graph at \(x = 6\) has an open dot at \(y=-2\) and maybe a different point? Wait, no, looking at the graph, at \(x = 6\), the open dot is at \(y=-2\), but is there a closed dot? Wait, the graph shows that at \(x = 6\), the function has an open dot, and when we check \(f(6)\), maybe the function's value at \(x = 6\) is not \(-2\)? Wait, no, actually, from the graph, when \(x = 6\), the open dot is at \(y=-2\), but maybe the function is defined elsewhere? Wait, no, the key is: \(\lim_{x
ightarrow6}f(x)=-2\) (since left and right limits are \(-2\)), but \(f(6)\): looking at the graph, at \(x = 6\), the open dot is at \(y = - 2\), but is there a closed dot? Wait, the graph has a point at \(x = 6\)? Wait, no, the user's graph: at \(x = 6\), there is an open dot (so the function is not continuous there in terms of the value, but the limit exists. Wait, but \(f(6)\): if we look at the graph, maybe \(f(6)\) is not equal to \(-2\)? Wait, no, maybe I made a mistake. Wait, the third statement is \(\lim_{x
ightarrow6}f(x)=f(6)\). Let's re - check. The limit as \(x
ightarrow6\) is \(-2\) (since left and right limits are \(-2\)). Now, what is \(f(6)\)? From the graph, at \(x = 6\), there is an open dot at \(y=-2\) and a closed dot? Wait, no, the graph shows that at \(x = 6\), the open dot is at \(y=-2\), but maybe the function's value at \(x = 6\) is not \(-2\). Wait, actually, in the graph, at \(x = 6\), the open dot is at \(y=-2\), but if we look at the graph, maybe \(f(6)\) is not defined as \(-2\). So \(\lim_{x
ightarrow6}f(x)=-2\), but \(f(6)
eq - 2\) (because of the open dot, maybe the function has a different value or is not defined there in the way that \(f(6)\) is not equal to the limit). So statement III is false.

Since statements I and II are true, the answer is C.

Answer:

C. I and II only