Problem 42 - Thomas Calculus

Question

Prove the limit statements in Exercises \(37-50\) $$ \lim _{x \rightarrow-2} f(x)=4 \quad \text { if } \quad f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \neq-2} \\ {1,} & {x=-2}\end{array}\right. $$

Step-by-Step Solution

Answer

The limit is 4 as \( x \to -2 \).

Step 1: Identify the Function and Limit

The function is piecewise: \[f(x) = \begin{cases} x^2, & \text{if } x eq -2 \ 1, & \text{if } x = -2 \end{cases}\]We are asked to find \( \lim_{x \to -2} f(x) \) which involves determining the behavior of \( f(x) \) as \( x \) approaches \(-2\) from both sides.

Step 2: Evaluate the Limit from the Left and Right

As \( x \) approaches \(-2\) from the left and right, \( x eq -2 \), so \( f(x) = x^2 \). Therefore, calculate:\[\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} x^2 = (-2)^2 = 4\]\[\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} x^2 = (-2)^2 = 4\]

Step 3: Check if Left and Right Limits are Equal

Since both the left-side limit and the right-side limit as \( x \) approaches \(-2\) are 4, the two-sided limit exists and is equal to 4. Thus,\[\lim_{x \to -2} f(x) = 4\]

Step 4: Confirm Continuity or Discontinuity at \(-2\)

Check the function value at \( x = -2 \), which is \( f(-2) = 1 \). Notice that \( f(-2) eq \lim_{x \to -2} f(x) \). This implies \( f(x) \) is not continuous at \( x = -2 \), but this does not affect the limit.

Key Concepts

Piecewise Functions

Understanding piecewise functions is crucial in calculus, especially when dealing with limits. A piecewise function is defined by different expressions based on the values or intervals of the input variable. In our exercise, the function \[f(x) = \begin{cases} x^2, & \text{if } x eq -2 \ 1, & \text{if } x = -2 \end{cases}\]is a piecewise function. This function behaves differently depending on whether or not \( x \) equals \(-2\).

• For \( x eq -2 \), \( f(x) \) is \( x^2 \).
• For \( x = -2 \), \( f(x) \) is \( 1 \).

The piecewise nature is pivotal because it tells us how to approach the function as \( x \) nears a given value, in this case, \(-2\). By focusing on \( x^2 \) for all \( x eq -2 \), we can determine the limits from either side of \(-2\). Piecewise functions allow for flexibility in defining functions that may have sudden changes or jumps at certain points.

Continuity

Continuity is an important concept in calculus that determines whether a function behaves smoothly without interruptions. A function is continuous at a point \( c \) if:

• The function is defined at \( c \); \( f(c) \) exists.
• The limit of the function as it approaches \( c \) exists.
• The value of the function at \( c \) is equal to the limit as \( x \) approaches \( c \).

In our scenario, we're specifically probing the continuity at \( x = -2 \). The function \( f(x) \) at this point changes to 1, but the limit as \( x \to -2 \) is 4. Since \( f(-2) eq \lim_{x \to -2} f(x) \), the function is discontinuous at \( x = -2 \). It's crucial to grasp that while a function may not be continuous at a point, it can still have a well-defined limit at that same point.

Left and Right Limits

Analyzing left and right limits helps us understand the behavior of a function as it approaches a particular point from either direction. A left limit is the value that a function approaches as the input approaches a certain point from the left (denoted \( x \to c^- \)), whereas the right limit approaches from the right (denoted \( x \to c^+ \)).
In our example:

• The left limit of \( f(x) \) as \( x \to -2^- \) is calculated using \( x^2 \), and equals 4.
• Similarly, the right limit of \( f(x) \) as \( x \to -2^+ \) also equals 4.

When these two one-sided limits are equal, i.e., \( \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = 4 \), the two-sided limit exists and is exactly that value. Hence, \( \lim_{x \to -2} f(x) = 4 \). Even if the function value at that point differs, the existence of equal left and right limits signifies smooth behavior approaching the point.