Problem 12 - Essential Calculus

Question

In Exercises \(11-14,\) find each limit, if possible. (a) \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x^{3}-1}\) (b) \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x-1}\) (c) \(\lim _{x \rightarrow \infty} \frac{3-2 x^{2}}{3 x-1}\)

Step-by-Step Solution

Answer

The solutions are: (a) 0, (b) -2/3, (c) undefined.

Step 1: Analyzing (a)

Let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x^{3}-1}\). If we plug \(x = \infty\) into the function, it will lead to an indeterminacy of the form 0/0. To resolve this indeterminacy, we will divide the numerator and the denominator by \(x^3\). So the function becomes \(\lim _{x \rightarrow \infty} \frac{3/x^{3}-2/x^{2}}{3 -1/x^{3}}\).

Step 2: Solving (a)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x^{3}\), \(2/x^{2}\), and \(1/x^{3}\) each approach 0 because they are constants divided by a large number. Therefore, the function becomes \( \lim _{x \rightarrow \infty} \frac{-2/x^{2}}{3}\), which equals to 0.

Step 3: Analyzing (b)

Now, let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x}{3 x-1}\). If we plug \(x = \infty\) into the function, it will yet again lead to an indeterminacy of the form 0/0. To resolve this, we will divide the numerator and the denominator by \(x\). So the function becomes \(\lim _{x \rightarrow \infty} \frac{3/x -2}{3 -1/x}\).

Step 4: Solving (b)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x\), and \(1/x\) each approach 0, hence, \( \lim _{x \rightarrow \infty} \frac{-2}{3}\), which equals to -2/3.

Step 5: Analyzing (c)

Finally, let's analyze the function \(\lim _{x \rightarrow \infty} \frac{3-2 x^{2}}{3 x-1}\). If we plug \(x = \infty\) into the function, it results in an indeterminacy of the form 0/0. To resolve this, we divide the numerator and the denominator by \(x^2\). The function becomes \(\lim _{x \rightarrow \infty} \frac{3/x^{2} -2}{3/x -1/x^{2}}\).

Step 6: Solving (c)

Now take the limit as x approaches infinity. As \(x\) approaches infinity, \(3/x^{2}\), \(3/x\), and \(1/x^{2}\) each approach 0, hence, \( \lim _{x \rightarrow \infty} \frac{-2}{0}\), which is undefined.

Key Concepts

Indeterminate Forms

When dealing with limits, you may encounter expressions that do not initially have a clear outcome. These are known as indeterminate forms. An indeterminate form arises when the direct substitution of a value into a limit results in an expression that does not provide enough information about the behavior of the function.

Some common types of indeterminate forms in calculus include:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(\infty - \infty\)
• \(0 \cdot \infty\)
• \(1^\infty\)
• \(0^0\)
• \(\infty^0\)

These forms suggest that the limit could converge to a specific number, diverge to infinity, or have no limit at all depending on how the functions grow or shrink relative to each other. In the exercise you provided, the forms \(\frac{3-2x}{3x^3-1}\), \(\frac{3-2x}{3x-1}\), and \(\frac{3-2x^2}{3x-1}\) are examples where \(\frac{0}{0}\) initially appears, necessitating further analysis to determine the limits.

Infinity in Calculus

Infinity often comes into play in calculus when analyzing the behavior of functions as they grow without bound, either positively or negatively. The symbol \(\infty\) is used to describe a value larger than any real number or, conversely, a value smaller than any negative real number, which we denote as \(-\infty\).

In the context of limits, infinity represents the idea that as \(x\) grows larger and larger (or as \(x\) becomes negative large), the function values might approach a certain behavior or pattern. Calculating the limit of a function as \(x\) approaches infinity helps us understand how the function behaves at these extreme values.

In the exercises you are working on, each function involves fractions where \(x\) tends toward infinity:

• For \(\lim_{x \rightarrow \infty} \frac{3 - 2x}{3x^3 - 1}\), the terms \(3/x^3\) and \(1/x^3\) vanish into insignificance as \(x\) grows very large, simplifying the expression.
• Similarly, in \(\lim_{x \rightarrow \infty} \frac{3 - 2x}{3x - 1}\), the terms \(3/x\) and \(1/x\) approach zero, giving/leading to a simple conclusion for the limit.
• Finally, in \(\lim_{x \rightarrow \infty} \frac{3 - 2x^2}{3x - 1}\), we encounter division by an increasingly small number which leads to the undefined behavior.

Understanding how expressions behave as they approach infinity is crucial in analyzing the full picture of how limits influence calculus.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus to resolve indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule provides a method for evaluating limits of indeterminate forms by differentiating the numerator and the denominator separately and then taking the limit of the resulting expression.

The rule states that for functions \(f(x)\) and \(g(x)\), if \(\lim_{{x \to c}} f(x) = 0\) and \(\lim_{{x \to c}} g(x) = 0\) or both approach infinity, and the derivatives \(f'(x)\) and \(g'(x)\) exist near \(c\), then:\[\lim_{{x \to c}} \frac{{f(x)}}{{g(x)}} = \lim_{{x \to c}} \frac{{f'(x)}}{{g'(x)}}\]provided the right-hand limit exists.

For the type of exercisex you worked on, while L'Hôpital's Rule could be applied to each initial indeterminate form, simplification techniques such as factorization or direct algebraic manipulation with limits provide a straightforward alternative. After algebraic simplification, unnecessary use of L'Hôpital's Rule is often avoided as demonstrated in the solution steps. Knowing when and how to use L'Hôpital's Rule can significantly simplify otherwise complicated limit problems, especially when initial simplification is not apparent.