Problem 78 - Calculus and its applications

Question

Find each limit, if it exists. If a limit does not exist, state that fact. a) \(\lim _{x \rightarrow 0} \frac{|x|}{x}\) b) \(\lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}-4}\)

Step-by-Step Solution

Answer

a) The limit does not exist. b) The limit is -3.

Step 1: Understanding Limit a

We need to evaluate the limit \(\lim _{x \rightarrow 0} \frac{|x|}{x}\). This expression involves the absolute value function and changes behavior around \(x = 0\).

Step 2: Analyzing Behavior from Left

As \(x\) approaches 0 from the left (\(x \to 0^-\)), \(|x| = -x\) because \(x\) is negative. Thus, \(\frac{|x|}{x} = \frac{-x}{x} = -1\).

Step 3: Analyzing Behavior from Right

As \(x\) approaches 0 from the right (\(x \to 0^+\)), \(|x| = x\) since \(x\) is positive. Thus, \(\frac{|x|}{x} = \frac{x}{x} = 1\).

Step 4: Conclusion for Limit a

Since the left-hand limit is \(-1\) and the right-hand limit is \(1\), the limits from both sides do not match. Therefore, \(\lim _{x \rightarrow 0} \frac{|x|}{x}\) does not exist.

Step 5: Understanding Limit b

Next, evaluate the limit \(\lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}-4}\). Notice that both the numerator and the denominator go to 0 as \(x = -2\), indicating a \(\frac{0}{0}\) indeterminate form.

Step 6: Factoring and Simplifying b

The numerator \(x^3 + 8\) can be factored as \((x+2)(x^2-2x+4)\) (sum of cubes), and the denominator \(x^2 - 4\) factors as \((x+2)(x-2)\).

Step 7: Cancel Common Factors

Cancel out the common factor \((x+2)\) from the numerator and the denominator, simplifying the expression to \(\frac{x^2-2x+4}{x-2}\).

Step 8: Evaluate the New Limit

Now evaluate the limit: \(\lim _{x \to -2} \frac{x^2-2x+4}{x-2}\). Plugging in \(x = -2\): \(\left((-2)^2 - 2(-2) + 4\right) = 4 +4 + 4 = 12\) and \(-2 - 2 = -4\). Thus, the limit is \(-\frac{12}{4} = -3\).

Step 9: Conclusion for Limit b

The limit for part b, \(\lim _{x \rightarrow-2} \frac{x^{3}+8}{x^{2}-4}\), is \(-3\).

Key Concepts

Absolute Value Function

The absolute value function, denoted as \(|x|\), essentially measures the distance of a number from zero on the number line, without considering direction. It is defined as follows: for any real number \(x\), \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\). This definition leads to some unique behaviors and properties, especially when evaluating limits.

• At points where the input changes sign, such as near zero, the function can behave differently based on whether the input is approaching from the positive or negative side.
• In the context of limits, understanding the behavior of the absolute value function is crucial to determining the limit accurately, especially when evaluating expressions like \[\lim _{x ightarrow 0} \frac{|x|}{x}\].

When approaching from the left, where \(x\) is negative, the expression simplifies as \(|x| = -x\), resulting in \(-1\). Whereas from the right, where \(x\) is positive, you have \(|x| = x\), resulting in \(1\). Because these two results do not match, the limit does not exist. It's a perfect example of how the absolute value function can affect limit calculations.

Indeterminate Form

An indeterminate form occurs in calculus when the limit of a function yields an indeterminate expression like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not provide straightforward answers, hence require further manipulation to determine a meaningful limit.

• Indeterminate forms indicate a need for additional mathematical techniques, such as factoring, rationalizing, or applying L'Hôpital's rule, to evaluate a limit.
• When you come across an expression in the form of \(\frac{0}{0}\), it's a sign that the function needs simplification or transformation before computation.

In the exercise, we observe \[\lim _{x ightarrow-2} \frac{x^{3}+8}{x^{2}-4}\], which presents a \(\frac{0}{0}\) indeterminate form. To resolve this, techniques like factoring the numerator and denominator were utilized to find the significant factors that allow simplification and cancellation, thus allowing the limit to be evaluated straightforwardly.

Factoring and Simplification

Factoring and simplification are powerful tools in calculus used to resolve limits that at first seem indeterminate or complex.

• Factoring allows breaking down complex expressions into simpler components, often revealing cancelable terms that result in a simplified expression.
• Simplification involves reducing expressions to their simplest form to make limits more manageable to calculate.

In the given example, the expression \(x^3 + 8\) was factored using the sum of cubes formula, resulting in \((x+2)(x^2-2x+4)\), and the denominator \(x^2 - 4\) was factored as \((x+2)(x-2)\). By canceling the common factor \(x+2\), the expression becomes \((x^2-2x+4)/(x-2)\). This creates a simpler expression to evaluate the limit at \(x = -2\), resulting in a straightforward calculation that avoids the indeterminacy found initially. Such simplifications are crucial in managing complex limit problems efficiently.