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How is \(\lim _{x \rightarrow a} p(x)\) calculated if \(p\) is a polynomial function?

Short Answer

Expert verified
Answer: The limit of a polynomial function as x approaches a specific value a is the value of the polynomial function at x=a.

Step by step solution

01

1. Understand polynomial functions

Polynomial functions can be defined as the sum of terms, each consisting of a constant (coefficient) multiplied by a power of the independent variable. The general form of a polynomial function is \(p(x) = c_nx^n + c_{n-1}x^{n-1} + ... + c_1x + c_0\), where \(c_i\) are the coefficients (constants) and \(n\) is a non-negative integer. Polynomial functions are continuous and differentiable everywhere.
02

2. Properties of limits

When solving limits, it's essential to know the properties associated with them. Some of these properties are: 1. The limit of a constant is the constant itself, i.e., \(\lim_{x\to a} c = c\). 2. The limit of the independent variable as it approaches some value is that value, i.e., \(\lim_{x\to a} x = a\). 3. The limit of a sum or difference of functions is the sum or difference of their respective limits, i.e., \(\lim_{x\to a} (f(x) \pm g(x)) = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x)\). 4. The limit of a product of functions is the product of their respective limits, i.e., \(\lim_{x\to a} (f(x) \cdot g(x)) = \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x)\).
03

3. Calculate the limit

Since polynomial functions are continuous and differentiable everywhere, we can find the limit by direct substitution, i.e., substitute \(x\) with \(a\) in the polynomial function \(p(x)\). Using the properties of limits mentioned previously, we have: \(\lim_{x\to a} p(x) = \lim_{x\to a} (c_nx^n + c_{n-1}x^{n-1} + ... + c_1x + c_0)\) Applying the properties, we get: \(\lim_{x\to a} p(x) = c_n ( \lim_{x\to a} x^n) + c_{n-1}( \lim_{x\to a} x^{n-1}) + ... + c_1( \lim_{x\to a} x) + c_0\) Now, using the limit of the independent variable, property 2: \(\lim_{x\to a} p(x) = c_na^n + c_{n-1}a^{n-1} + ... + c_1a + c_0\) Hence, the limit of a polynomial function as \(x\) approaches \(a\) is the value of the polynomial at \(x=a\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Function
A polynomial function is a mathematical expression made up of a sum of terms, where each term consists of a constant (called a coefficient) multiplied by a power of the variable (usually \(x\)). The general form of a polynomial function is written as:\[ p(x) = c_n x^n + c_{n-1} x^{n-1} + \, ... \, + c_1 x + c_0 \]Here:
  • \(c_n, c_{n-1}, ..., c_1, c_0\) are constants known as coefficients.
  • \(n\) is a non-negative integer, indicating the highest power of the polynomial, which is also called the degree of the polynomial.
Polynomial functions are very valuable because of their properties. They are continuous and differentiable everywhere on the real number line. This means they do not have breaks, jumps, or sharp turns, and you can find their derivative easily. Understanding polynomial functions is crucial for evaluating limits, solving equations, and in many applications in science and engineering.
Properties of Limits
When we need to calculate the limit of a function, we use several useful properties of limits. Understanding these properties makes solving limit problems much more manageable. Here are some essential properties:
  • The limit of a constant is the constant itself: \( \lim_{x \to a} c = c \).
  • The limit of the variable \(x\) as it approaches some value \(a\) is \(a\): \( \lim_{x \to a} x = a \).
  • The limit of a sum or difference of functions is equal to the sum or difference of their respective limits: \( \lim_{x \to a} (f(x) \pm g(x)) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \).
  • The limit of a product of functions is the product of their limits: \( \lim_{x \to a} (f(x) \cdot g(x)) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \).
By utilizing these properties, you can break down complex limit problems into simpler parts, which can be evaluated more straightforwardly. This makes it especially helpful when dealing with polynomial functions, where these properties allow us to evaluate their limits efficiently.
Direct Substitution
Direct substitution is a straightforward technique used to find the limit of a function, especially when dealing with polynomial functions. Because polynomial functions are continuous, you can find the limit by simply substituting the given value directly into the polynomial. Here’s how it works:For a polynomial \( p(x) = c_n x^n + c_{n-1} x^{n-1} + \, ... \, + c_1 x + c_0 \), the limit as \( x \to a \) is:\[ \lim_{x \to a} p(x) = c_n a^n + c_{n-1} a^{n-1} + \, ... \, + c_1 a + c_0 \]The reason this works so effortlessly is due to the continuity of polynomials. Since polynomial functions don’t have any breaks or undefined points, the value they take at \( x = a \) is exactly the limit value as \( x \to a \).To practically apply direct substitution:
  • Take the given polynomial.
  • Replace \(x\) with \(a\) in the expression.
  • Simplify the expression to get the value of the limit.
This technique is very efficient and eliminates the guesswork often involved with more complex functions where direct substitution cannot be used.

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Most popular questions from this chapter

Suppose \(\$ 5000\) is invested in a savings account for 10 years ( 120 months), with an annual interest rate of \(r\) compounded monthly. The amount of money in the account after 10 years is given by \(A(r)=5000(1+r / 12)^{120}\) a. Show there is a value of \(r\) in \((0,0.08)-\) an interest rate between \(0 \%\) and \(8 \%-\) that allows you to reach your savings goal of \(\$ 7000\) in 10 years. b. Use a graph to illustrate your explanation in part (a). Then approximate the interest rate required to reach your goal.

Determine whether the following statements are true and give an explanation or counterexample. a. If a function is left-continuous and right-continuous at \(a\), then it is continuous at \(a\) b. If a function is continuous at \(a\), then it is left-continuous and right- continuous at \(a\) c. If \(a< b\) and \(f(a) \leq L \leq f(b),\) then there is some value of \(c\) in \((a, b)\) for which \(f(c)=L\) d. Suppose \(f\) is continuous on \([a, b] .\) Then there is a point \(c\) in \((a, b)\) such that \(f(c)=(f(a)+f(b)) / 2\)

One-sided limits a. Evaluate \(\lim _{x \rightarrow 2^{+}} \sqrt{x-2}\). b. Explain why \(\lim _{x \rightarrow 2^{-}} \sqrt{x-2}\) does not exist.

Evaluate \(\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}\).

Assume you invest \(\$ 250\) at the end of each year for 10 years at an annual interest rate of \(r .\) The amount of money in your account after 10 years is given by \(A(r)=\frac{250\left((1+r)^{10}-1\right)}{r} .\) Assume your goal is to have \(\$ 3500\) in your account after 10 years. a. Show that there is an interest rate \(r\) in the interval \((0.01,0.10)-\) between \(1 \%\) and \(10 \%-\) that allows you to reach your financial goal. b. Use a calculator to estimate the interest rate required to reach your financial goal.

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