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Chapter 1: Problem 4

Find the limits. $$\lim _{x \rightarrow 3} x^{3}-3 x^{2}+9 x$$

Short Answer

Expert verified
The limit is 27.

Step by step solution

01

Identify the Function

The function given is a polynomial function: \( f(x) = x^3 - 3x^2 + 9x \). We need to find the limit of this function as \( x \) approaches 3.
02

Apply the Limit Directly

For polynomial functions, the limit can be found by direct substitution, as polynomials are continuous everywhere. Substitute \( x = 3 \) into the function: \[ f(3) = 3^3 - 3(3)^2 + 9(3) \]
03

Calculate the Expression

Calculate the value \( f(3) \) using the substituted value: \[ 3^3 = 27 \] \[ 3(3)^2 = 3 \times 9 = 27 \] \[ 9 \times 3 = 27 \] So, \[ f(3) = 27 - 27 + 27 \]
04

Simplify the Result

Simplify the expression obtained in the previous step: \[ f(3) = 27 - 27 + 27 = 27 \]
05

Conclude the Limit

Thus, the limit \( \lim_{x \to 3} (x^3 - 3x^2 + 9x) \) is \( 27 \).

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Key Concepts

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

Polynomial Functions
A polynomial function consists of terms with variable powers that are non-negative integers, and each term is formed by multiplying a numerical coefficient by a variable to a power. The general form of a polynomial can be written as: \[ a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]where \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are constants and \( x \) is the variable.
  • In our exercise, the polynomial is \( x^3 - 3x^2 + 9x \).
  • Each part of the polynomial is a monomial, with the first term being cubic, the second quadratic, and the third linear.
Polynomials are fundamental because they are simple and entirely determined by their degree and coefficients. They play a significant role in calculus since their properties make them easy to work with.
Continuous Functions
A function is continuous at a point if there is no interruption in the graph of the function at that point. To be more specific, a function \( f(x) \) is continuous at \( x = a \) if the following three conditions are met:
  • \( f(a) \) is defined.
  • The limit \( \lim_{x \to a} f(x) \) exists.
  • The limit \( \lim_{x \to a} f(x) = f(a) \).
For polynomial functions, like \( x^3 - 3x^2 + 9x \), continuity is a given because polynomials are smooth and continuous everywhere on their domain, which is all real numbers (-鈭�, 鈭�). This is why they are useful for limit evaluation, as there are no breaks or holes to consider around any point. This property helps us examine limits using simple techniques like direct substitution, knowing the function behaves nicely all around the point of interest.
Direct Substitution Method
The direct substitution method is used to find the limit of a function by simply replacing the variable \( x \) with the value that \( x \) is approaching. This method is very handy for functions that are continuous at the point you are examining.
  • For the given exercise, we apply this method because the function \( x^3 - 3x^2 + 9x \) is continuous at \( x = 3 \).
  • By substituting \( x = 3 \) into the function, we directly compute \( 3^3 - 3 \times 3^2 + 9 \times 3 \).
After performing the substitution and calculations, we find the outcome to be \( 27 \). The direct substitution method simplifies finding limits in continuous functions, making it a straightforward and powerful approach in calculus.

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Most popular questions from this chapter

Find the fallacy in the following "proof" that \(\frac{1}{8}>\frac{1}{4}\) Multiply both sides of the inequality \(3>2\) by \(\log \frac{1}{2}\) to $$\begin{aligned}\text { get } & 3 \log \frac{1}{2}>2 \log \frac{1}{2} \\\\\log \left(\frac{1}{2}\right)^{3} &>\log \left(\frac{1}{2}\right)^{2} \\\\\log \frac{1}{8} &>\log \frac{1}{4} \\ \frac{1}{8} &>\frac{1}{4}\end{aligned}$$

Prove: (a) \(\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \quad(|x|<1)\). (b) \(\cos ^{-1} x=\frac{\pi}{2}-\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}} \quad(|x|<1)\).

Determine whether the statement is true or false, Explain your answer. The graph of the exponential function with base \(b\) passes through the point (0,1)

Prove: If \(p(x)\) is a polynomial of odd degree, then the equation \(p(x)=0\) has at least one real solution.

(a) Use the Intermediate-Value Theorem to show that the equation \(x=\cos x\) has at least one solution in the interval \([0, \pi / 2]\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.
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