Problem 31 - Thomas Calculus

Question

Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{x \rightarrow 1} \frac{x^{-1}-1}{x-1}$$

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Recognize the Indeterminate Form

When substituting the value of the limit directly, i.e., when x = 1, the expression becomes \(\frac{1^{-1} - 1}{1 - 1}\), which simplifies to \(\frac{0}{0}\). This indicates an indeterminate form, so further analysis is needed, such as algebraic manipulation or finding another approach to evaluate the limit.

Step 2: Apply L'Hopital's Rule

Since the limit is initially in an indeterminate form of \(\frac{0}{0}\), we can apply L'Hopital's Rule. This rule states that for limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), the limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) is equal to \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\) if this second limit exists. Differentiate the numerator and the denominator separately.

Step 3: Differentiate the Numerator and Denominator

The numerator is \(x^{-1} - 1\). Its derivative, using the power rule for derivatives, is \(-x^{-2}\). The denominator is \(x - 1\), and its derivative is \(1\).

Step 4: Apply Derived Functions in L'Hopital's Rule

Substitute the derivatives back into the limit expression:\[\lim_{x \to 1} \frac{-x^{-2}}{1} = \lim_{x \to 1} -x^{-2}.\]

Step 5: Evaluate the Limit

Now, substitute \(x = 1\) into the simplified expression:\[\lim_{x \to 1} -x^{-2} = -1^{-2} = -1.\]

Step 6: State the Result

The limit of the original expression as \(x\) approaches \(1\) is \(-1\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool in calculus, especially when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). If you encounter such forms while trying to evaluate a limit, L'Hôpital's Rule can often resolve the issue.

Here's how it works:

• After confirming that the limit is in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), you can switch from working with the original functions \(f(x)\) and \(g(x)\) to their derivatives \(f'(x)\) and \(g'(x)\).
• The expression \(\lim_{x \to c} \frac{f(x)}{g(x)}\) becomes \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\).
• The key is that this new limit should be simpler to evaluate and, importantly, exist.

This method is powerful because it transforms the problem of dealing with tricky quotients into one of simpler derivatives. You perform simple calculations on these new functions, and they lead you to the correct limit.

Indeterminate Forms

Indeterminate forms signal uncertainty or indeterminacy in limit problems. These forms occur when substitution of the limit value into the function initially yields results like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which don't give any clear information about the limit.

Let's break these down:

• \(\frac{0}{0}\) suggests both the numerator and denominator become zero at the limit point; hence, there's no actual division defined.
• \(\frac{\infty}{\infty}\) indicates both parts are growing without bound, yet how they compare is unclear.

Indeterminate forms require further analysis. Without additional steps, these forms can’t directly determine a function’s limit. Techniques like algebraic manipulation, factoring, rationalizing, or applying L'Hôpital's Rule help find the actual limit where these forms appear.

Derivatives

Derivatives are a fundamental part of calculus, representing the rate of change of a function with respect to a variable. When you differentiate a function, you essentially find how the function behaves locally.

For instance:

• The derivative of a power function, like \(x^{-1}\), is found using the power rule: bring the exponent down and reduce the exponent by one. So, the derivative of \(x^{-1}\) is \(-x^{-2}\).
• For a linear function, like \(x - 1\), the derivative is straightforward: any constant term becomes zero, and the variable term retains its coefficient.

In applying L'Hôpital's Rule, you'll often calculate these derivatives for both the numerator and the denominator. This step simplifies the limit expression and makes it possible to evaluate it more easily.