Problem 50 - Precalculus

Question

Evaluate the limits using the limit properties. $$\lim _{x \rightarrow-8}\left(\frac{1}{2} x^{2}-\sqrt[3]{x}+3\right)$$

Step-by-Step Solution

Answer

The limit is 33.

Step 1: Identify the Limit Expression

We are given the expression \(\frac{1}{2} x^{2} - \sqrt[3]{x} + 3\) and asked to find its limit as \(x\) approaches \(-8\). This expression comprises a quadratic term, a cube root term, and a constant.

Step 2: Apply Limit Properties Separately

We can split the overall limit into the sum of limits for each term:\[\lim_{x \to -8} \left( \frac{1}{2} x^2 - \sqrt[3]{x} + 3 \right) = \lim_{x \to -8} \frac{1}{2} x^2 - \lim_{x \to -8} \sqrt[3]{x} + \lim_{x \to -8} 3\]Each of these simpler limits can be evaluated independently.

Step 3: Evaluate the Quadratic Term Limit

Calculate the limit of the quadratic term:\[\lim_{x \to -8} \frac{1}{2} x^2 = \frac{1}{2} \times (-8)^2 = \frac{1}{2} \times 64 = 32\]This term evaluates to 32 as \(x\) approaches \(-8\).

Step 4: Evaluate the Cube Root Term Limit

Calculate the limit of the cube root term:\[\lim_{x \to -8} \sqrt[3]{x} = \sqrt[3]{-8} = -2\]This term evaluates to \(-2\) as \(x\) approaches \(-8\).

Step 5: Evaluate the Constant Term Limit

The limit of a constant is simply the constant itself, so:\[\lim_{x \to -8} 3 = 3\]

Step 6: Combine All Limits

Substitute the independently evaluated limits back into the sum equation:\[\lim_{x \to -8} \left( \frac{1}{2} x^2 - \sqrt[3]{x} + 3 \right) = 32 - 2 + 3 = 33\]

Step 7: Conclusion

The limit of \( \frac{1}{2} x^{2} - \sqrt[3]{x} + 3 \) as \(x\) approaches \(-8\) is 33.

Key Concepts

Limit Properties

When evaluating limits, breaking them down into smaller, manageable pieces can make the process simpler. This is where limit properties come in handy. For an expression with multiple terms, like \( \frac{1}{2} x^{2} - \sqrt[3]{x} + 3 \), we use the rule that allows us to separate limits for addition and subtraction.

• If you have \( \lim_{x \to a} [f(x) + g(x)] \), you can split it as \( \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \).
• This property also works for subtraction.

By doing this, we handle each component of the expression independently, simplifying the overall task.

This principle not only applies to sums and differences but also makes evaluating products and quotients straightforward under certain conditions.

Quadratic Term

The quadratic term in our expression is \( \frac{1}{2} x^2 \). Let's explore what happens to this term as \( x \) approaches \(-8\). Quadratic terms involve squared variables, which means the variable is raised to the power of two.

• When evaluating \( \lim_{x \to -8} \frac{1}{2} x^2 \), we substitute \( x = -8 \) into the expression.
• This gives us \( \frac{1}{2} \times (-8)^2 \).
• Calculate the square: \((-8)^2 = 64\).

Now multiply: \( \frac{1}{2} \times 64 = 32 \).
Thus, the limit of the quadratic term is 32 as \( x \) approaches \(-8\). Quadratic terms always yield positive results when squared, which helps in anticipating the behavior of limits.

Cube Root Term

The cube root term, \( \sqrt[3]{x} \), behaves differently from quadratic terms because we're looking to identify the number that, when multiplied by itself three times, equals the given number.
For \( x = -8 \), the limit calculation becomes straightforward:

• First, consider \( \lim_{x \to -8} \sqrt[3]{x} \).
• Finding the cube root of \(-8\) means finding a number that cubed equals \(-8\).
• That number is \(-2\), since \((-2)^3 = -8\).

Therefore, the limit of the cube root term is \(-2\) as \( x \) approaches \(-8\). Cube roots handle negative and positive numbers differently than square roots, offering a negative result when applied to negative numbers.

Constant Term

The constant term in our expression is the simplest of all.
The term \(3\) stays unchanged regardless of the value of \( x \).

• The limit \( \lim_{x \to a} c \) for any constant \( c \) is simply \( c \).
• Thus, \( \lim_{x \to -8} 3 = 3 \).

Constant terms do not vary with \( x \), so their limits remain the same, no matter where \( x \) approaches on the number line. This makes them a reliable and consistent factor in expressions involving limits.