Question

Use a calculator to graph the function and estimate the value of the limit, then use L'Hôpital's rule to find the limit directly. $$ \lim _{x \rightarrow 1} \frac{e^{(x-1)}-1}{x-1} $$

Short answer

The limit is 1.

Step-by-step solution

Graph the Function on a Calculator

Use a graphing calculator to plot the function \( y = \frac{e^{(x-1)}-1}{x-1} \). Observe the behavior of the graph as \( x \) approaches 1. The graph will help provide a visual estimate of the limit.

Visual Estimation from the Graph

From the graph, as \( x \) approaches 1, the value of \( y \) appears to approach a certain finite value. Make a note of this estimated value, but remember that this needs to be verified using calculus.

Recognize the Indeterminate Form

Evaluate the function directly at \( x = 1 \). Substituting \( x = 1 \) into the equation gives \( \frac{e^{(1-1)}-1}{1-1} = \frac{0}{0} \), which is an indeterminate form necessitating the use of L'Hôpital's Rule.

Apply L'Hôpital's Rule

Identify that L'Hôpital's Rule applies since the limit is in the \( \frac{0}{0} \) form. Differentiate the numerator and the denominator separately: - The derivative of \( e^{(x-1)}-1 \) is \( e^{(x-1)} \).- The derivative of \( x-1 \) is 1.Use these derivatives to set up the new limit expression.

Evaluate the Limit Using L'Hôpital's Rule

Now, rewrite and evaluate the limit: \[ \lim_{x \to 1} \frac{e^{(x-1)}}{1} = \lim_{x \to 1} e^{(x-1)} \].Substitute \( x = 1 \) into the new expression to calculate: \( e^{(1-1)} = e^0 = 1 \).

Conclude the Solution

With the visual estimation and the application of L'Hôpital's Rule, it's confirmed the limit is 1. Hence, \( \lim_{x \to 1} \frac{e^{(x-1)}-1}{x-1} = 1 \).

Key concepts

Understanding Indeterminate Forms

When dealing with limits in calculus, you may encounter expressions that don't initially have a clear result. These are called "indeterminate forms." A common example is the form \( \frac{0}{0} \), which occurs in the given limit problem \( \lim_{x \to 1} \frac{e^{(x-1)}-1}{x-1} \). This happens because both the numerator and the denominator approach zero as \( x \) approaches 1. Unlike simple zero, results, these forms require special techniques, like L'Hôpital's Rule, to resolve. Indeterminate forms signal that further analysis using calculus is necessary to determine the actual value of the limit.

Using a Graphing Calculator

A graphing calculator is a powerful tool for visualizing mathematical functions. When you graph the function \( y = \frac{e^{(x-1)} - 1}{x-1} \), you can visually estimate the behavior of the function as \( x \) nears a specific value, in this case, 1. Here are some tips for using a graphing calculator effectively:

• Input the function accurately to ensure the correct graph is displayed.
• Observe the curve as it approaches \( x = 1 \). Seeing the graph level out or approach a specific height gives an initial estimated value for the limit.
• Remember that the visual estimation is just that—an estimate and should be verified using analytical methods like calculus.

Utilizing a graphing calculator in tandem provides a good mix of visual insight and mathematical accuracy.

Derivatives and Their Importance

Derivatives are fundamental in calculus and play a crucial role in evaluating limits, particularly when using L'Hôpital's Rule. A derivative represents the rate at which a function is changing at any point along its graph. In our exercise, we calculate the derivatives needed for L'Hôpital's procedure:

• The derivative of \( e^{(x-1)} - 1 \) is \( e^{(x-1)} \), reflecting how the function grows or shrinks near each point.
• The derivative of \( x - 1 \) is simply 1, as this is a linear function with a constant slope.

By evaluating these derivatives, you transform the original indeterminate form into something manageable, allowing straightforward limit evaluation.

Mastering Limit Evaluation

Limit evaluation is a critical skill in calculus where you determine the behavior of functions as variables approach specific values. With L'Hôpital's Rule, it's possible to resolve limits presenting as indeterminate forms, such as \( \frac{0}{0} \). Here's how you execute limit evaluation using L'Hôpital's Rule:

• Begin by assessing whether the given limit results in an indeterminate form.
• Differentiate both the numerator and denominators separately to rewrite the limit expression using these derivatives.
• Re-evaluate the limit with the simplified expression. In this case, \( \lim_{x \to 1} e^{(x-1)} \), directly substituting \( x = 1 \) yields \( e^{0} = 1 \).

Mastering this process ensures precise and reliable conclusions for complex limit problems, enhancing your overall calculus proficiency.