Problem 20 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}$$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Verify Indeterminate Form

First, substitute \(x = 0\) into the expression to see if it results in an indeterminate form. The function is \(\frac{e^{x}-e^{-x}-2 x}{x- ext{sin} x}\). At \(x = 0\), the numerator becomes \(e^{0}-e^{0}-2(0) = 1 - 1 = 0\) and the denominator becomes \(0 - \sin(0) = 0 - 0 = 0\). Since we have a \(\frac{0}{0}\) form, l'Hospital's Rule can be applied.

Step 2: Apply l'Hospital's Rule

Apply l'Hospital's Rule, which involves taking the derivatives of the numerator and the denominator separately. The derivative of the numerator, \(e^{x} - e^{-x} - 2x\), is \(e^{x} + e^{-x} - 2\). The derivative of the denominator, \(x - \sin x\), is \(1 - \cos x\). So, the limit becomes \(\lim_{x \to 0} \frac{e^{x} + e^{-x} - 2}{1 - \cos x}\).

Step 3: Evaluate the New Limit

Substitute \(x = 0\) again in the new expression \(\frac{e^{x} + e^{-x} - 2}{1 - \cos x}\). The numerator becomes \(e^{0} + e^{0} - 2 = 1 + 1 - 2 = 0\) and the denominator becomes \(1 - \cos(0) = 1 - 1 = 0\). The form \(\frac{0}{0}\) still exists; l'Hospital's Rule must be applied again.

Step 4: Apply l'Hospital's Rule Again

Take the second derivatives of the numerator and the denominator. The new derivative of the numerator \(e^{x} + e^{-x} - 2\) is \(e^{x} - e^{-x}\). The new derivative of the denominator \(1 - \cos x\) is \(\sin x\). The limit now becomes \(\lim_{x \to 0} \frac{e^{x} - e^{-x}}{\sin x}\).

Step 5: Evaluate the Limit Again

Substitute \(x = 0\) into the new expression \(\frac{e^{x} - e^{-x}}{\sin x}\). The numerator becomes \(e^{0} - e^{0} = 1 - 1 = 0\) and the denominator becomes \(\sin(0) = 0\). We again encounter \(\frac{0}{0}\) and must apply l'Hospital's Rule once more.

Step 6: Apply l'Hospital's Rule a Third Time

Take the third derivatives: The new derivative of the numerator \(e^{x} - e^{-x}\) is \(e^{x} + e^{-x}\). The derivative of \(\sin x\) is \(\cos x\). The limit now becomes \(\lim_{x \to 0} \frac{e^{x} + e^{-x}}{\cos x}\).

Step 7: Final Evaluation

Substitute \(x = 0\) into the latest expression. The numerator becomes \(e^{0} + e^{0} = 1 + 1 = 2\) and the denominator becomes \(\cos(0) = 1\). Thus, the result is \(\frac{2}{1}\).

Key Concepts

Limit Evaluation

When you're tackling calculus problems, evaluating limits is often crucial. The limit of a function as it approaches a specific value gives insight into the function's behavior at that point. Limits can appear complicated, but the core idea is to understand how close the function's values get to a specific number as the input gets closer to a particular point.

In the example provided, we are asked to find the limit as \( x \) approaches zero for the expression \( \frac{e^x - e^{-x} - 2x}{x - \sin x} \). Initially, we substitute \( x = 0 \) into the expression; however, we encounter an undefined situation, or indeterminate form, which we'll explore further in the next section. Thus, direct substitution wasn't enough, and further steps are necessary to evaluate it properly through means like l'Hospital's Rule or possibly other algebraic simplifications.

Indeterminate Form

An indeterminate form occurs in calculus when evaluating a function does not initially give a clear result. Classic types include \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms require us to apply special techniques, such as l'Hospital's Rule, to resolve them.

• In our example, using \( x = 0 \), the attempt to calculate \( \frac{e^x - e^{-x} - 2x}{x - \sin x} \) leads to \( \frac{0}{0} \).
• This indeterminate result indicates the need for other methods, since just plugging in numbers fails to find a usable limit.

Indeterminate forms don't provide straightforward answers, and that's why calculus techniques that rely on derivatives become essential tools in finding limits, such as l'Hospital's Rule, which we frequently apply in this problem.

Derivatives

Derivatives measure how a function changes as its input changes. They are fundamental in limit problems, especially when using l'Hospital's Rule. This rule involves taking the derivatives of both the numerator and the denominator of a fraction separately.

In the exercise, l'Hospital's Rule was applied three separate times:

• First, we derived \( e^x + e^{-x} - 2 \) over \( 1 - \cos x \).
• Then, \( e^x - e^{-x} \) over \( \sin x \).
• Finally, \( e^x + e^{-x} \) over \( \cos x \).

The ability to compute derivatives like these swiftly is necessary, as they directly impact the ease and speed with which you can resolve limits, especially those initially problematic like our indeterminate forms.

Exponential Functions

Exponential functions are a crucial topic in calculus, often appearing in limit evaluation problems like our example. An exponential function is of the form \( e^x \), and it’s characterized by the constant base \( e \), approximately 2.718.

• The behavior of exponential functions is smooth and continuous, with crucial properties aiding in finding limits.
• In our exercise, both \( e^x \) and \( e^{-x} \) feature prominently. These functions grow rapidly as \( x \) increases, and their derivatives \( e^x \) and \( -e^{-x} \) reflect their sensitive nature to changes in \( x \).

Understanding these functions' properties simplifies working with them in calculus, especially for applying methods like l'Hospital's Rule to resolve limit problems.