Problem 85 - Calculus and its applications

Question

Fill in each blank so that \(\lim _{x \rightarrow 2} f(x)\) exists. $$ f(x)=\left\\{\begin{array}{ll} x^{2}-9, & \text { for } x<2, \\ -x^{2}+\ldots, & \text { for } x>2 \end{array}\right. $$

Step-by-Step Solution

Answer

The value of \(a\) must be \(-1\) for the limit as \(x\) approaches 2 to exist.

Step 1: Identify the Limit Condition

To ensure that \(\lim _{x \rightarrow 2} f(x)\) exists, \(f(x)\) must approach the same value from both the left and right as \(x\) approaches 2. This means that the limit from the left, \(\lim_{x \to 2^-} f(x)\), and the limit from the right, \(\lim_{x \to 2^+} f(x)\), must be equal.

Step 2: Calculate the Left-Hand Limit

For \(x < 2\), the function is defined as \(f(x) = x^2 - 9\). Calculate \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 - 9) = 2^2 - 9 = 4 - 9 = -5\).

Step 3: Set Up the Right-Hand Limit

For \(x > 2\), the function is \(f(x) = -x^2 + a\), where \(a\) is the constant we need to find. We need \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-x^2 + a)\) to match the left limit of \(-5\).

Step 4: Calculate and Equate the Right-Hand Limit

Calculate \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-x^2 + a) = -(2^2) + a = -4 + a\). Set this equal to \(-5\): \[-4 + a = -5\].

Step 5: Solve for \(a\)

Solve the equation \(-4 + a = -5\) to find \(a\). Add \(4\) to both sides: \[a = -1\].

Key Concepts

Left-Hand Limit

When we talk about the left-hand limit, we're interested in what happens to a function as we approach a specific point from the left side. Think about zooming in on a number line from values that are slightly less than 2. The left-hand limit is represented as \( \lim_{x \to c^-} f(x) \), where \( c \) is the point you're approaching. When dealing with piecewise functions like the one in the original exercise, ensure that you use the correct part of the function for calculating the left-hand approach.
For \( x < 2 \), we use the equation \( f(x) = x^2 - 9 \). When calculating the limit as \( x \) approaches 2 from the left, plug in values closer to 2, like 1.9, and notice how \( f(x) = 1.9^2 - 9 \) trends toward a single number.
In our example,

• Calculate \( \lim_{x \to 2^-} (x^2 - 9) \).
• Substitute and simplify: \( 2^2 - 9 = -5 \).

The left-hand limit is \(-5\). This value is crucial as it should match the right-hand limit for the function's overall limit to exist.

Right-Hand Limit

The right-hand limit pertains to the function's behavior as we approach a specified point from the right. Imagine closing in on a number on the right, like sliding down the number line towards it from a bit more than your target. This is represented mathematically as \( \lim_{x \to c^+} f(x) \).

In our original exercise, for values of \( x > 2 \), we use the part of the function equation \( f(x) = -x^2 + a \). When you evaluate the right-hand limit, plug in values just greater than 2, such as 2.1, to see how these values dictate the trend toward a single number.
To make sure the function's overall limit exists, you should analyze and set the right-hand limit:

• Calculate \( \lim_{x \to 2^+} (-x^2 + a) \).
• Substitute and simplify: \( -(2^2) + a = -3 + a \).

You need the result to match the left-hand limit found earlier, which was \(-5\). Therefore:

• Set \( -4 + a = -5 \).
• Solving gives \( a = -1 \).

Existence of Limits

The concept of the existence of limits is fundamental in calculus. For a limit to exist at a specific point, both the left-hand limit and right-hand limit should approach the same value. If both sides zero in on the same number as we approach the point, that's when we say \( \lim_{x \to c} f(x) \) exists.
In our problem, the exercise's aim was to determine the limit at where \( x \to 2 \). This means that both:

• The left-hand limit from step 2: \( -5 \)
• The right-hand limit, calculated with value for \( a \) found in step 4: \(-5 \)

When these conditions are satisfied, the overall limit \( \lim_{x \to 2} f(x) \) is established to exist. This critical analysis of both directions of approach ensures function consistency and confirms that you've found the correct configuration of your piecewise function.