Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-28, use a calculator to find the indicated limit. Use a graphing calculator to plot the function near the limit point. $$ \lim _{t \rightarrow 1} \frac{t^{2}-1}{\sin (t-1)} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Understand the Problem

The problem asks us to find the limit of the function \( \frac{t^2-1}{\sin(t-1)} \) as \( t \) approaches 1.

Step 2: Simplify the Expression

Recognize that \( t^2 - 1 = (t-1)(t+1) \). Thus, the expression becomes \( \frac{(t-1)(t+1)}{\sin(t-1)} \).

Step 3: Apply Limit Laws

Since both the numerator \( (t-1)(t+1) \) and the denominator \( \sin(t-1) \) approach zero as \( t \) approaches 1, we have an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule.

Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator separately. The derivative of \((t-1)(t+1)\) is \(2t\), and the derivative of \(\sin(t-1)\) is \(\cos(t-1)\). Thus, we have the new expression \( \lim _{t \rightarrow 1} \frac{2t}{\cos(t-1)} \).

Step 5: Evaluate the Limit

Substitute \( t = 1 \) into the new expression: \( \frac{2 \times 1}{\cos(1-1)} = \frac{2}{\cos(0)} = \frac{2}{1} = 2 \).

Step 6: Verify with a Graphing Calculator

Plot the function \( \frac{t^2-1}{\sin(t-1)} \) on the graphing calculator near \( t = 1 \) to visually confirm that the function approaches 2 as \( t \) approaches 1.

Key Concepts

Limits

Limits in calculus are used to define the value that a function approaches as the input approaches some point. They are fundamental in analyzing behaviors at points of interest within a function, especially where direct substitution is not feasible.
Imagine a situation where you're attempting to determine the value of a function at a certain point where the function itself is not directly defined. That’s where limits shine. For instance, when asked for the limit \[\lim_{t \rightarrow 1} \frac{t^2 - 1}{\sin(t-1)}\] you're essentially observing the function's behavior as the variable 't' heads towards 1.

• Limits allow us to handle cases where functions approach certain behaviors, like approaching zero or infinity.
• They are critical in finding derivatives, which are essentially limits themselves.
• Understanding limits helps in studying continuous functions and determining where discontinuities may occur.

As seen in your exercise, identifying expressions that yield indeterminate forms, such as \(\frac{0}{0}\), is a crucial part of employing limits effectively.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for resolving indeterminate forms, particularly \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). When you encounter these forms while trying to compute a limit, direct substitution won’t suffice, but L'Hôpital's Rule saves the day!
This rule allows you to differentiate both the numerator and the denominator separately and then take the limit of that new fraction. Let's explore how it works in your problem.
We started with: \[\lim_{t \rightarrow 1} \frac{(t-1)(t+1)}{\sin(t-1) }\] Here, substituting directly gives \(\frac{0}{0}\), a classic indeterminate form. Applying L'Hôpital's Rule, differentiate the top and bottom separately to obtain:

• The derivative of \((t-1)(t+1)\) results in \(2t\).
• The derivative of \(\sin(t-1)\) becomes \(\cos(t-1)\).

So, L'Hôpital's Rule restructures our limit problem into: \[\lim_{t \rightarrow 1} \frac{2t}{\cos(t-1)}\]
Evaluating at \(t = 1\) gives us an answer of 2. With this rule, resolving indeterminate forms becomes significantly more approachable, rather than plain substitution, which often doesn’t work.

Indeterminate Forms

Indeterminate forms appear in limits when direct computation yields undefined expressions, like \(\frac{0}{0}\) or \(\infty - \infty\). In calculus, recognizing these forms is vital because they signal the need for additional mathematical techniques to find a solution.
When dealing with limits, expressions might sometimes result in shapes that aren't straightforwardly solvable. If we look at \[ \lim_{t \rightarrow 1} \frac{t^2-1}{\sin(t-1)} \] a thorough substitution gives us \(\frac{0}{0}\). That’s a textbook example of an indeterminate form, prompting us to use strategies like L'Hôpital's Rule or algebraic simplification.

• Encountering \(\frac{0}{0}\) often involves evaluating derivatives if L'Hôpital's Rule applies.
• Other forms like \(0 \times \infty\) or \(1^{\infty}\) may require different approaches, including logarithmic transformations or factorization.
• Understanding that an expression is indeterminate guides you towards using the right tools to untangle it.

Recognizing these forms leads to developing a sense of which method or rule might successfully unravel the problem at hand, often requiring additional insight into the function's behavior around the limit of interest.