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Chapter 2: Problem 23

Exercise \(1-26:\) Find the limit, if it exists. $$ \lim _{x \rightarrow 2^{3}} \cdot \frac{x^{2}}{4-9 x^{2}} $$

Short Answer

Expert verified
The limit is \( \frac{16}{-143} \).

Step by step solution

01

Substitute the Limit into the Function

First, we substitute the value of the limit into the function directly to see if it results in a determinate form. Here, we have:\[ \lim_{x \to 8} \frac{x^2}{4 - 9x^2} \]Substituting \( x = 8 \), the function becomes:\[ \frac{8^2}{4 - 9 \cdot 8^2} = \frac{64}{4 - 576} = \frac{64}{-572} \]
02

Simplify the Expression

In this step, we simplify the expression obtained from substituting the limit. The numerator is 64, and the denominator is -572, which can be simplified by dividing both by their greatest common divisor.\[ \frac{64}{-572} = \frac{16}{-143} \]
03

Verify the Result

Ensure that the substitution results in a determinate value and not an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Since substituting directly gave us a valid fraction, this confirms that the limit exists and has been correctly calculated.

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Key Concepts

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

Substitution Method
The substitution method is a straightforward approach to finding limits in calculus. It's about directly injecting the value the variable is approaching into the given function.
Often, this first attempt provides a clear answer. Here's how it works step-by-step:
  • Identify the value for which the limit is to be calculated. In our example, the limit as \( x \) approaches 8.
  • Substitute this value directly into the function.
  • Calculate the result. The goal is to see if it results in a determinate number, which means no indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
When you can directly replace the variable and arrive at a valid number, like \( \frac{64}{-572} \) in this case, it indicates that the limit exists. This method is effective when the function behaves nicely, meaning there's no division by zero or other undefined operations at the point you're examining. It's always the first step you should try when tackling a limit problem.
Fraction Simplification
Simplifying fractions is an essential skill in mathematics, and it's especially useful when dealing with limits. After substitution, you'll often end up with a fraction that can be simplified to make working with it easier.
Simplifying a fraction means expressing it in the smallest possible terms.
  • Find the greatest common divisor (GCD) of the numerator and the denominator.
  • Divide both the numerator and the denominator by this GCD.
In our exercise, after substitution, we arrive at \( \frac{64}{-572} \). By dividing both numbers by their GCD, which is 4, the fraction simplifies to \( \frac{16}{-143} \). This simplification doesn't change the value but makes any further calculations less cumbersome and reduces the risk of error.
Determinate Form
In calculus, determinate and indeterminate forms are huge when it comes to limits. A determinate form means you鈥檝e got a solid answer, not some undefined or problematic expression.
When you calculate a limit and substitute the variable, the result needs to be clean:
  • You should avoid forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), which are considered indeterminate because they don't give clear answers.
  • If you reach a determinate form鈥攋ust a plain fraction or a whole number鈥攊t's your goal!
In our example, when we substituted \( x = 8 \) into the expression, we achieved \( \frac{64}{-572} \), a determinate form. This signifies the limit exists and is properly calculated. Remember, verifying for a determinate form confirms the integrity of your solution, indicating that no mathematical anomaly was involved in your substitution and simplification processes.

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Most popular questions from this chapter

A function \(f\) satisfies the given conditions. Sketch a possible graph for \(f\), assuming that it does not cross a horizontal asymptote. $$ \lim _{x \rightarrow-x} f(x)=1 ; \quad \lim _{x \rightarrow \infty} f(x)=1 $$

Find all numbers at which \(f\) is continuous. $$ f(x)=\frac{|x+9|}{x+9} $$

Exercise \(1-26:\) Find the limit, if it exists. $$ \lim _{x \rightarrow 3} \frac{1}{5 x-3} $$

Exercise \(35-38:\) Find all numbers at which \(f\) is discontinuous. $$ f(x)=\frac{1}{x^{2}-16} $$

Find each limit, if it exists: |a) \(\lim _{x \rightarrow a^{-}} f(x)\) (b) \(\lim _{x \rightarrow a^{+}} f(x)\) (c) \(\lim _{x \rightarrow a} f(x)\) $$ f(x)=\frac{|x-4|}{x-4}: \quad a=4 $$
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