Question

practice exercises 19–26. evaluating limits graphically sketch a graph of f and use it to make a conjecture about the values of f(a), (limlimits_{x \to a^-} f(x)), (limlimits_{x \to a^+} f(x)), and (limlimits_{x \to a} f(x)) or state that they do not exist. 19. ( f(x) = \begin{cases} x^2 + 1 & \text{if } x leq -1 \\ 3 & \text{if } x > -1 end{cases}; a = -1 ) 20. ( f(x) = \begin{cases} 3 - x & \text{if } x < 2 \\ x - 1 & \text{if } x > 2 end{cases}; a = 2 ) 21. ( f(x) = \begin{cases} sqrt{x} & \text{if } x < 4 \\ 3 & \text{if } x = 4 \\ x + 1 & \text{if } x > 4 end{cases}; a = 4 )

Explanation:

Problem 19:

Step 1: Find \( f(-1) \)

For \( x = -1 \), we use the first part of the piecewise function (\( x \leq -1 \)): \( f(-1) = (-1)^2 + 1 = 1 + 1 = 2 \).

Step 2: Find \( \lim_{x \to -1^-} f(x) \)

As \( x \) approaches \( -1 \) from the left (\( x < -1 \)), we use \( f(x) = x^2 + 1 \). So, \( \lim_{x \to -1^-} (x^2 + 1) = (-1)^2 + 1 = 2 \).

Step 3: Find \( \lim_{x \to -1^+} f(x) \)

As \( x \) approaches \( -1 \) from the right (\( x > -1 \)), we use \( f(x) = 3 \). So, \( \lim_{x \to -1^+} 3 = 3 \).

Step 4: Find \( \lim_{x \to -1} f(x) \)

Since \( \lim_{x \to -1^-} f(x) = 2 \) and \( \lim_{x \to -1^+} f(x) = 3 \), and these two limits are not equal, \( \lim_{x \to -1} f(x) \) does not exist.

Step 1: Find \( f(2) \)

The piecewise function does not define \( f(2) \) (since \( x < 2 \) and \( x > 2 \) are given, but not \( x = 2 \)), so \( f(2) \) is undefined.

Step 2: Find \( \lim_{x \to 2^-} f(x) \)

As \( x \) approaches \( 2 \) from the left (\( x < 2 \)), we use \( f(x) = 3 - x \). So, \( \lim_{x \to 2^-} (3 - x) = 3 - 2 = 1 \).

Step 3: Find \( \lim_{x \to 2^+} f(x) \)

As \( x \) approaches \( 2 \) from the right (\( x > 2 \)), we use \( f(x) = x - 1 \). So, \( \lim_{x \to 2^+} (x - 1) = 2 - 1 = 1 \).

Step 4: Find \( \lim_{x \to 2} f(x) \)

Since \( \lim_{x \to 2^-} f(x) = 1 \) and \( \lim_{x \to 2^+} f(x) = 1 \), the two-sided limit exists and is \( 1 \).

Step 1: Find \( f(4) \)

For \( x = 4 \), we use the second part of the piecewise function: \( f(4) = 3 \).

Step 2: Find \( \lim_{x \to 4^-} f(x) \)

As \( x \) approaches \( 4 \) from the left (\( x < 4 \)), we use \( f(x) = \sqrt{x} \). So, \( \lim_{x \to 4^-} \sqrt{x} = \sqrt{4} = 2 \).

Step 3: Find \( \lim_{x \to 4^+} f(x) \)

As \( x \) approaches \( 4 \) from the right (\( x > 4 \)), we use \( f(x) = x + 1 \). So, \( \lim_{x \to 4^+} (x + 1) = 4 + 1 = 5 \).

Step 4: Find \( \lim_{x \to 4} f(x) \)

Since \( \lim_{x \to 4^-} f(x) = 2 \) and \( \lim_{x \to 4^+} f(x) = 5 \), and these two limits are not equal, \( \lim_{x \to 4} f(x) \) does not exist.

Answer:

\( f(-1) = 2 \), \( \lim_{x \to -1^-} f(x) = 2 \), \( \lim_{x \to -1^+} f(x) = 3 \), \( \lim_{x \to -1} f(x) \) does not exist.

Problem 20: