Problem 17 - Calculus: Early Transcendentals

Question

Suppose \(p\) and \(q\) are polynomials. If \(\lim _{x \rightarrow 0} \frac{p(x)}{q(x)}=10\) and \(q(0)=2,\) find \(p(0)\).

Step-by-Step Solution

Answer

Answer: The value of p(0) is 20.

Step 1: Write the limit equation

Given that, \(\lim _{x \rightarrow 0}\frac{p(x)}{q(x)}=10\), we need to use this equation to find the relationship between p(x) and q(x) as x approaches 0.

Step 2: Use the property of limits

We can rewrite the given limit using the property of limits for a fraction: \(\lim _{x \rightarrow 0} \frac{p(x)}{q(x)}= \frac{\lim _{x \rightarrow 0} p(x)}{\lim _{x \rightarrow 0} q(x)}\) Then, using the given information, we can substitute the limit and q(0) values: \(10 = \frac{\lim _{x \rightarrow 0} p(x)}{2}\)

Step 3: Solve for \(\lim _{x \rightarrow 0} p(x)\)

Now we can solve the equation for \(\lim _{x \rightarrow 0} p(x)\): \(\lim _{x \rightarrow 0} p(x) = 10 \cdot 2\) \(\lim _{x \rightarrow 0} p(x) = 20\) Since \(q(x)\) is continuous at \(x = 0\) (because \(q(0)\) exists), we can conclude that \(p(x)\) must also be continuous at this point. That means we can substitute \(p(0)\) for \(\lim _{x \rightarrow 0} p(x)\): \(p(0) = 20\) So, the value of p(0) is 20.

Key Concepts

Polynomials

Polynomials are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. A polynomial in variable \(x\) looks like \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0\), where \(n\) is a non-negative integer and \(a_n, a_{n-1}, \ldots, a_0\) are constants called coefficients.

Understanding polynomials is crucial because they often appear in calculus-related problems. They are simple yet versatile, serving as a foundation for more complex functions. With polynomials, you can easily apply calculus techniques like differentiation and integration. They also have much predictable properties that simplify analysis:

• They are defined for all real numbers; there's no worry about division by zero or undefined square roots.
• Their behavior as \(x\) becomes large is controlled by the leading term \(a_nx^n\).

In solving the given problem, both \(p(x)\) and \(q(x)\) are polynomials. This means their limits as \(x\) approaches 0 are straightforward because their values can be accessed directly by substitution if continuous.

Continuous Functions

Continuous functions are those where small changes in the input result in small changes in the output. Mathematically, a function \(f(x)\) is continuous at a point \(a\) if \(\lim_{x \to a} f(x) = f(a)\).

In practical terms, if you can draw a curve without lifting your pencil, you're likely dealing with a continuous function. Polynomials, like \(p(x)\) and \(q(x)\) in the problem, are continuous everywhere on their domain. This continuity allows us to directly compute limits by simply substituting the point into the polynomial.

For our exercise, since \(q(x)\) is continuous at \(x = 0\) and \(q(0) = 2\), we were able to use the known limit properties and given conditions effectively to solve for \(p(0)\). In continuous functions, as exhibited here, substituting the point to evaluate a limit is accurate and straightforward.

Limit Properties

Limits are a fundamental concept in calculus used to solve problems involving behavior as a function approaches a specific point. The property essential here is that if you have a limit of a quotient as \(x\) approaches a value, it can be expressed as the limit of the numerator over the limit of the denominator, provided the denominator does not become zero.

For functions \(p(x)\) and \(q(x)\), the limit equation was \(\lim _{x \rightarrow 0} \frac{p(x)}{q(x)} = 10\). This equation indicates that as \(x\) approaches 0, the behavior of the fraction is equivalent to a constant 10.

The steps in the solution used these limit properties to rewrite the fraction's limit, allowing the seamless application of the known limits of the numerator and denominator. We replaced the limits with their respective values: since \(\lim_{x \to 0} q(x) = q(0) = 2\), we deduced \(\lim_{x \to 0} p(x)\) through simple algebra, multiplying both sides by the known denominator. These properties are indispensable tools for efficiently solving limit problems across calculus.