Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\lim _{x \rightarrow 3} \frac{x^{3}+x-30}{x-3}\)

Step-by-Step Solution

Answer

The limit is 28.

Step 1: Identify the indeterminate form

First, we need to substitute the value of \( x = 3 \) into the given expression to check if it results in an indeterminate form. Substitute \( x = 3 \) into \( \frac{x^{3}+x-30}{x-3} \). We get: \( \frac{3^{3}+3-30}{3-3} = \frac{27+3-30}{0} = \frac{0}{0} \), which is indeed an indeterminate form. Therefore, we will need to simplify the expression.

Step 2: Simplify the numerator

The numerator \( x^3 + x - 30 \) must be factorized to eliminate the indeterminate form. Since this is a cubic polynomial, apply polynomial long division or try factoring it directly. We can test for possible factors by substituting various values into the equation or apply synthetic division. Let's try dividing \( x^3 + x - 30 \) by \( x - 3 \).

Step 3: Factor using synthetic division

Using synthetic division, divide \( x^3 + x - 30 \) by \( x - 3 \). Set up the synthetic division with the value \( 3 \). Coefficients are \( 1, 0, 1, -30 \). After performing the synthetic division, we get a quotient of \( x^2 + 3x + 10 \) with a remainder of \(0\). Thus, \( x^3 + x - 30 = (x - 3)(x^2 + 3x + 10) \).

Step 4: Rewrite the expression and find the limit

Now that the numerator is factorized, the expression becomes \( \frac{(x - 3)(x^2 + 3x + 10)}{x - 3} \). Cancel the \( x - 3 \) terms from the numerator and the denominator: \( x^2 + 3x + 10 \). Finally, substitute \( x = 3 \) in the simplified polynomial: \( 3^2 + 3 \times 3 + 10 = 9 + 9 + 10 = 28 \).

Step 5: Conclusion

After cancelling the common factor in the indeterminate form and substituting \( x = 3 \) into the simplified expression, the limit is calculated to be \( 28 \). Thus, the limit of the original function as \( x \) approaches \( 3 \) is \( 28 \).

Key Concepts

Limits

In calculus, limits help us understand how a function behaves as it approaches a specific point. A limit is the value that a function or sequence "approaches" as the input or index approaches some value. They are crucial in calculus because they lay the foundation for derivatives and integrals.
To find the limit of a function, we can often substitute the point we're approaching into the function. However, sometimes direct substitution results in an undefined expression, and we need other techniques to evaluate the limit. For instance, in the exercise \[ \lim _{x \rightarrow 3} \frac{x^{3}+x-30}{x-3} \]simply plugging in \( x = 3 \) would've given us an undefined form. Hence, we explore additional methods such as factorization or division to find the limit.

Indeterminate Forms

An indeterminate form often appears in calculus when evaluating limits. One such form occurs when directly substituting a number into a function results in \( \frac{0}{0} \). This is not a valid result, because dividing by zero is undefined. To resolve this, we must simplify or transform the expression to find the limit.
In our exercise, substituting \( x = 3 \) into \( \frac{x^3 + x - 30}{x-3} \) gives an indeterminate form: \( \frac{0}{0} \). This indicates that more work is needed to simplify the function or apply another method, like factoring, before continuing with finding the limit.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, particularly useful when the divisor is a linear factor. Instead of the lengthy polynomial long division, synthetic division uses only the coefficients of the polynomial, simplifying calculations significantly.
In the given example, we used synthetic division to divide \( x^3 + x - 30 \) by \( x - 3 \). The process involves setting up the division using the root of the divisor (in this case, \( x = 3 \)), and performing quick arithmetic to find the quotient and remainder. In this instance, we found that the remainder was zero, confirming that \( x - 3 \) is a factor of the polynomial, leading to further simplification.

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as a product of its factors. This can simplify expression evaluations and solve polynomial equations. Factoring is especially useful in calculus for simplifying expressions when finding limits or solving derivative problems.
In the exercise, after identifying that \( x^3 + x - 30 \) had a factor of \( x - 3 \) using synthetic division, we factorized the polynomial as \( (x - 3)(x^2 + 3x + 10) \). This factorization allowed us to cancel out the common term in the numerator and denominator, removing the indeterminate form and making it easier to evaluate the limit.