Question

the graph below is the function $f(x)$
determine which one of the following rules for continuity is first violated at $a = -1$.
$\bigcirc$ $f(a)$ is defined.
$\bigcirc$ $\lim\limits_{x \to a} f(x)$ exists.
$\bigcirc$ $\lim\limits_{x \to a} f(x) = f(a)$.
question help: video

Explanation:

Step1: Recall continuity rules

A function \( f(x) \) is continuous at \( x = a \) if three conditions are met:

1. \( f(a) \) is defined.
2. \( \lim_{x \to a} f(x) \) exists.
3. \( \lim_{x \to a} f(x) = f(a) \).

We check each condition at \( a = -1 \).

Step2: Check \( f(a) \) is defined

From the graph, at \( x = -1 \), there is a filled dot at \( y = 1 \), so \( f(-1) = 1 \). Thus, \( f(a) \) is defined.

Step3: Check \( \lim_{x \to a} f(x) \) exists

To find the limit as \( x \to -1 \), we check the left - hand limit (LHL) and right - hand limit (RHL).

• Left - hand limit: As \( x \) approaches \( -1 \) from the left, the function approaches \( - 3\) (since there is an open circle at \( (-1,-3) \) for the left - hand part of the graph).
• Right - hand limit: As \( x \) approaches \( -1 \) from the right, the function approaches \( 1\) (from the line \( y=-x \), when \( x = - 1\), \( y = 1\)).

Since LHL \(=-3\) and RHL \( = 1\), and LHL\(
eq\) RHL, \( \lim_{x\to - 1}f(x)\) does not exist.

Step4: Analyze the third condition (though not needed as we found the violation in step 3)

Since the limit does not exist, we don't need to check the third condition as the second condition for continuity is already violated.

Answer:

\(\lim\limits_{x\to a}f(x)\) exists.