Problem 26 - Calculus Single Variable

Question

Use an algebraic manipulation to put the limit in a form which can be treated using l'Hôpital's Rule; then evaluate the limit. \(\lim _{x \rightarrow 0}\left(e^{x}-e^{-x}\right) \cdot x^{-1}\)

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Express the Limit

The given limit is \( \lim _{x \rightarrow 0}\left(e^{x}-e^{-x}\right) \cdot x^{-1} \). To attempt to apply l'Hôpital's Rule, we first recognize that as \( x \rightarrow 0 \), the expression \( \frac{e^{x} - e^{-x}}{x} \) appears, suggesting a \( \frac{0}{0} \) form. Thus, the limit is currently indeterminate and suitable for l'Hôpital's Rule.

Step 2: Differentiate Numerator and Denominator

Apply l'Hôpital's Rule by differentiating the numerator and denominator separately. The derivative of the numerator, \( e^{x} - e^{-x} \), is \( e^{x} + e^{-x} \). The derivative of the denominator, \( x \), is 1. Thus, the new limit expression is \( \lim_{x \rightarrow 0} \frac{e^{x} + e^{-x}}{1} \).

Step 3: Evaluate the Simplified Limit

With the new expression \( \lim_{x \rightarrow 0}(e^{x} + e^{-x}) \), substitute \( x = 0 \) directly into the expression: \( e^{0} + e^{0} = 1 + 1 = 2 \). Therefore, the limit evaluates to 2.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms arise in limit problems when substituting the limiting value directly into the expression does not yield a clear result. This typically happens with forms such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0\cdot\infty \), among others. These forms indicate that the direct substitution will not work, and another method must be applied to find the limit.In our example, we are dealing with the limit \( \lim _{x \rightarrow 0}\left(e^{x}-e^{-x}\right) \cdot x^{-1} \). As \( x \rightarrow 0 \), both the numerator \( e^x - e^{-x} \) and the denominator \( x \) approach zero, creating a \( \frac{0}{0} \) indeterminate form. Recognizing this form tells us that we need a strategy like l'Hôpital's Rule to evaluate the limit properly.

Algebraic Manipulation

Algebraic manipulation is crucial in transforming mathematical expressions in a way that makes them easier to analyze. It involves rearranging or rewriting terms so the expression becomes more manageable or applicable to a certain rule, like l'Hôpital's Rule.In the given exercise, the expression \( \left(e^{x}-e^{-x}\right) \cdot x^{-1} \) is first rewritten as a fraction: \( \frac{e^{x} - e^{-x}}{x} \). This step is essential to recognize the indeterminate form and decide on the subsequent solution method. By rewriting the initial product as a fraction, we're preparing the expression for differentiation, allowing l'Hôpital's Rule to be applied effectively.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as the variable gets arbitrarily close to a certain point. When faced with indeterminate forms, direct evaluation doesn't work, so we have to use rules like l'Hôpital's Rule.In our problem, after applying algebraic manipulation and recognizing the \( \frac{0}{0} \) form, we apply l'Hôpital’s Rule. This rule involves differentiating the numerator and denominator separately to get a new limit to evaluate. Here, the differentiation gives us a new function, \( \frac{e^{x} + e^{-x}}{1} \). Simplifying this, the task of limit evaluation becomes straightforward. Evaluating as \( x \rightarrow 0 \), we substitute and find \( e^{0} + e^{0} = 2 \). Thus, the final evaluated limit is 2.