Problem 43 - APEX Calculus

Question

Let \(f(x)=\left\\{\begin{array}{cc}x^{2}-5 & x<5 \\ 5 x & x \geq 5\end{array}\right.\). (a) \(\lim _{x \rightarrow 5^{-}} f(x)\) (c) \(\lim _{x \rightarrow 5} f(x)\) (b) \(\lim _{x \rightarrow 5^{+}} f(x)\) (d) \(f(5)\)

Step-by-Step Solution

Answer

(a) 20; (b) 25; (c) Does not exist; (d) 25.

Step 1: Understanding the Function

The function \( f(x) \) is defined differently for values of \( x \) less than 5 and for values of \( x \) greater than or equal to 5. It is a piecewise function: \( f(x) = x^2 - 5 \) when \( x < 5 \), and \( f(x) = 5x \) when \( x \geq 5 \).

Step 2: Finding \( \lim_{x \to 5^-} f(x) \)

To find the left-hand limit as \( x \to 5 \) from the left, we consider \( f(x) = x^2 - 5 \). Substitute \( x = 5 \) into this expression to find the limit: \[ \lim_{x \to 5^-} f(x) = 5^2 - 5 = 25 - 5 = 20. \]

Step 3: Finding \( \lim_{x \to 5^+} f(x) \)

To find the right-hand limit as \( x \to 5 \) from the right, we use \( f(x) = 5x \). Substitute \( x = 5 \) into this expression: \[ \lim_{x \to 5^+} f(x) = 5 \times 5 = 25. \]

Step 4: Finding \( \lim_{x \to 5} f(x) \)

The full limit \( \lim_{x \to 5} f(x) \) exists only if the left-hand limit and right-hand limit are equal. Since \( \lim_{x \to 5^-} f(x) = 20 \) and \( \lim_{x \to 5^+} f(x) = 25 \), the limit does not exist because these two values are not equal.

Step 5: Evaluating \( f(5) \)

To find the value of the function at \( x = 5 \), we use the definition for \( x \geq 5 \, (5x) \). Substitute \( x = 5 \): \[ f(5) = 5 \times 5 = 25. \]

Key Concepts

Left-Hand Limit

When we talk about the left-hand limit, we imagine approaching a specific point from the left side on the function graph. For our piecewise function, we want to determine what happens as we move towards 5 from values less than 5. This is represented by the notation \( \lim_{x \to 5^-} f(x) \).
In this particular function, for \( x < 5 \), the expression \( f(x) = x^2 - 5 \) is used. If we substitute \( x = 5 \) into this equation to see what value the function approaches as we get very close from the left, we calculate:

• \( 5^2 - 5 = 25 - 5 = 20 \)

Thus, the left-hand limit as we approach 5 is 20.

Right-Hand Limit

The right-hand limit considers what happens to the function as we approach a specific point from the right side. This is indicated by \( \lim_{x \to 5^+} f(x) \). For the given function, for \( x \geq 5 \), the expression is \( f(x) = 5x \).
To find out what happens when \( x \) approaches 5 from values greater than or equal to 5, we substitute into the formula for \( x \geq 5 \):

• \( 5 \times 5 = 25 \)

This means the right-hand limit as \( x \) approaches 5 is 25.

Existence of Limits

The existence of limits is an important concept because it tells us whether the function approaches a single value as \( x \) approaches a certain point. The key requirement is that both the left-hand and right-hand limits need to be the same.
For this piecewise function at \( x = 5 \):

• The left-hand limit \( \lim_{x \to 5^-} f(x) = 20 \)
• The right-hand limit \( \lim_{x \to 5^+} f(x) = 25 \)

Since these two limits are not equal, the overall limit \( \lim_{x \to 5} f(x) \) does not exist. A unified value is necessary for a limit to exist at a given point.

Continuity

Continuity of a function at a certain point is observed when there are no jumps or gaps at that point. A function is continuous at a point if the limit as \( x \) approaches the point equals the function's value at that point, additionally ensuring both left-hand and right-hand limits are equal.
For our piecewise function:

• We found \( \lim_{x \to 5} f(x) \) does not exist due to unequal left and right limits.
• However, the value of the function at 5 is \( f(5) = 25 \) as determined by the expression \( f(x) = 5x \) for \( x \geq 5 \).

Since the overall limit doesn’t exist, the function is not continuous at \( x = 5 \). Continuity requires these elements to align perfectly without disparities.