Question

1. a) find the limit as ( x ) approaches 2 from the left, the limit as ( x ) approaches 2 from the right, and state whether ( lim_{x \to 2} f(x) ) exists.

b) find the limit as ( x ) approaches 1 from the left, the limit as ( x ) approaches 1 from the right, and state whether ( lim_{x \to 1} f(x) ) exists.

c) find the limit as ( x ) approaches 3 from the left, the limit as ( x ) approaches 3 from the right, and state whether ( lim_{x \to 3} f(x) ) exists.

d) find the limit as ( x ) approaches 4 from the left, the limit as ( x ) approaches 4 from the right, and state whether ( lim_{x \to 4} f(x) ) exists.

e) for which of the points ( x = 1, 2, 3, 4 ) is the function ( f ) continuous? explain why and why not they are continuous.

1. find the limit: ( lim \frac{x - 9}{} ) (incomplete part, but main focus on 16 with the graph above)

Explanation:

Step1: Analyze Left-Hand Limit (x→2⁻)

To find the limit as \( x \) approaches 2 from the left (\( x \to 2^- \)), observe the \( y \)-values of the graph as \( x \) gets closer to 2 from values less than 2. From the graph, as \( x \) approaches 2 from the left, the \( y \)-values approach 2. So, \( \lim_{x \to 2^-} f(x) = 2 \).

Step2: Analyze Right-Hand Limit (x→2⁺)

For the limit as \( x \) approaches 2 from the right (\( x \to 2^+ \)), look at the \( y \)-values as \( x \) gets closer to 2 from values greater than 2. The graph shows that as \( x \) approaches 2 from the right, the \( y \)-values approach 2. Thus, \( \lim_{x \to 2^+} f(x) = 2 \).

Step3: Determine Existence of \( \lim_{x \to 2} f(x) \)

Since the left-hand limit (\( \lim_{x \to 2^-} f(x) = 2 \)) and the right-hand limit (\( \lim_{x \to 2^+} f(x) = 2 \)) are equal, the two - sided limit \( \lim_{x \to 2} f(x) \) exists and is equal to 2.

Step4: Analyze Left-Hand Limit (x→1⁻)

For \( x \to 1^- \), observe the graph as \( x \) approaches 1 from values less than 1. The \( y \)-values approach 1. So, \( \lim_{x \to 1^-} f(x) = 1 \).

Step5: Analyze Right-Hand Limit (x→1⁺)

As \( x \) approaches 1 from the right (\( x \to 1^+ \)), the \( y \)-values from the graph approach 2. So, \( \lim_{x \to 1^+} f(x) = 2 \).

Step6: Determine Existence of \( \lim_{x \to 1} f(x) \)

Since \( \lim_{x \to 1^-} f(x)=1 \) and \( \lim_{x \to 1^+} f(x) = 2 \), and \( 1 eq2 \), the two - sided limit \( \lim_{x \to 1} f(x) \) does not exist.

Step7: Analyze Left-Hand Limit (x→3⁻)

For \( x \to 3^- \), looking at the graph, as \( x \) approaches 3 from values less than 3, the \( y \)-values approach 4. So, \( \lim_{x \to 3^-} f(x)=4 \).

Step8: Analyze Right-Hand Limit (x→3⁺)

As \( x \) approaches 3 from the right (\( x \to 3^+ \)), the \( y \)-values from the graph approach 3. So, \( \lim_{x \to 3^+} f(x)=3 \).

Step9: Determine Existence of \( \lim_{x \to 3} f(x) \)

Since \( \lim_{x \to 3^-} f(x) = 4 \) and \( \lim_{x \to 3^+} f(x)=3 \), and \( 4 eq3 \), the two - sided limit \( \lim_{x \to 3} f(x) \) does not exist.

Step10: Analyze Left-Hand Limit (x→4⁻)

For \( x \to 4^- \), as \( x \) approaches 4 from values less than 4, the \( y \)-values from the graph approach 3. So, \( \lim_{x \to 4^-} f(x)=3 \).

Step11: Analyze Right-Hand Limit (x→4⁺)

As \( x \) approaches 4 from the right (\( x \to 4^+ \)), the \( y \)-values from the graph approach 3. So, \( \lim_{x \to 4^+} f(x)=3 \).

Step12: Determine Existence of \( \lim_{x \to 4} f(x) \)

Since \( \lim_{x \to 4^-} f(x)=3 \) and \( \lim_{x \to 4^+} f(x)=3 \), the two - sided limit \( \lim_{x \to 4} f(x) \) exists and is equal to 3.

Step13: Analyze Continuity at \( x = 1,2,3,4 \)

- **At \( x = 1 \)**:
The function value \( f(1) \) (from the graph, the point at \( x = 1 \)) is 1. But \( \lim_{x \to 1} f(x) \) does not exist (since left - hand limit \( = 1 \) and right - hand limit \( = 2 \)). Also, \( \lim_{x \to 1} f(x) eq f(1) \) (even if we ignore the limit existence, \( 1 eq2\) and the limit doesn't exist). So, \( f(x) \) is not continuous at \( x = 1 \).
- **At \( x = 2 \)**:
We know \( \lim_{x \to 2} f(x)=2 \). From the graph, the function value \( f(2) \) is 2 (the point at \( x = 2 \)). Since \( \lim_{x \to 2} f(x)=f(2) \), the function is continuous at \( x = 2 \).
- **At \( x = 3 \)**:
The function value \( f(3) \) (from the graph, the point at \( x = 3 \)) is 4. But \( \lim_{x \to 3} f(x) \) does not exist (left - hand limit \( = 4 \), right - hand limit \( = 3 \)). Also, \( \lim_{x \to 3} f(x) eq f(3) \) (even if we ignore the limit existence, \( 3 eq4\) and the limit doesn't exist). So, \( f(x) \) is not continuous at \( x = 3 \).
- **At \( x = 4 \)**:
We know \( \lim_{x \to 4} f(x)=3 \). From the graph, the function value \( f(4) \) is 3 (the point at \( x = 4 \)). Since \( \lim_{x \to 4} f(x)=f(4) \), the function is continuous at \( x = 4 \).

Part (a)

\( \lim_{x \to 2^-} f(x)=2 \), \( \lim_{x \to 2^+} f(x)=2 \), \( \lim_{x \to 2} f(x)=2 \)

Part (b)

\( \lim_{x \to 1^-} f(x)=1 \), \( \lim_{x \to 1^+} f(x)=2 \), \( \lim_{x \to 1} f(x) \) does not exist

Part (c)

\( \lim_{x \to 3^-} f(x)=4 \), \( \lim_{x \to 3^+} f(x)=3 \), \( \lim_{x \to 3} f(x) \) does not exist

Part (d)

\( \lim_{x \to 4^-} f(x)=3 \), \( \lim_{x \to 4^+} f(x)=3 \), \( \lim_{x \to 4} f(x)=3 \)

Part (e)

- At \( x = 1 \): Not continuous (limit does not exist and \( \lim_{x\to1}f(x) eq f(1) \))
- At \( x = 2 \): Continuous ( \( \lim_{x\to2}f(x)=f(2) \))
- At \( x = 3 \): Not continuous (limit does not exist and \( \lim_{x\to3}f(x) eq f(3) \))
- At \( x = 4 \): Continuous ( \( \lim_{x\to4}f(x)=f(4) \))

For 17 (assuming the limit is \( \lim_{x\to a}\frac{x - y}{...} \) (the problem is incomplete, but if we assume a common limit like \( \lim_{x\to0}\frac{\sin x}{x}=1 \) or other standard limits, but since the problem is not fully shown, we can't solve it completely. If it's \( \lim_{x\to0}\frac{x - \sin x}{x^3} \), we can use L'Hopital's Rule:

First, check the form: as \( x\to0 \), \( \frac{0 - 0}{0}=\frac{0}{0} \), so we can apply L'Hopital's Rule.

First derivative of numerator: \( 1-\cos x \), derivative of denominator: \( 3x^2 \). Now we have \( \lim_{x\to0}\frac{1 - \cos x}{3x^2} \). This is still \( \frac{0}{0} \) (since \( 1-\cos(0)=0 \) and \( 3\times0^2 = 0 \)).

Second derivative of numerator: \( \sin x \), derivative of denominator: \( 6x \). Now we have \( \lim_{x\to0}\frac{\sin x}{6x} \). This is still \( \frac{0}{0} \) (since \( \sin(0)=0 \) and \( 6\times0 = 0 \)).

Third derivative of numerator: \( \cos x \), derivative of denominator: \( 6 \). Now we have \( \lim_{x\to0}\frac{\cos x}{6}=\frac{\cos(0)}{6}=\frac{1}{6} \). But this is based on an assumption about the limit. If the limit is different, the solution will change.

Answer:

Part (a)

\( \lim_{x \to 2^-} f(x)=\boldsymbol{2} \), \( \lim_{x \to 2^+} f(x)=\boldsymbol{2} \), \( \lim_{x \to 2} f(x)=\boldsymbol{2} \)

Part (b)

\( \lim_{x \to 1^-} f(x)=\boldsymbol{1} \), \( \lim_{x \to 1^+} f(x)=\boldsymbol{2} \), \( \lim_{x \to 1} f(x) \) does not exist

Part (c)

\( \lim_{x \to 3^-} f(x)=\boldsymbol{4} \), \( \lim_{x \to 3^+} f(x)=\boldsymbol{3} \), \( \lim_{x \to 3} f(x) \) does not exist

Part (d)

\( \lim_{x \to 4^-} f(x)=\boldsymbol{3} \), \( \lim_{x \to 4^+} f(x)=\boldsymbol{3} \), \( \lim_{x \to 4} f(x)=\boldsymbol{3} \)

Part (e)

• \( x = 1 \): Not continuous
• \( x = 2 \): Continuous
• \( x = 3 \): Not continuous
• \( x = 4 \): Continuous

For 17 (assuming \( \lim_{x\to0}\frac{x - \sin x}{x^3} \)):

\( \lim_{x\to0}\frac{x - \sin x}{x^3}=\boldsymbol{\frac{1}{6}} \) (Note: This is based on an assumption about the limit. If the limit is different, the answer will change)