Problem 48 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)^{2}}{x \sin x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Evaluate the Indeterminate Form

First, substitute \( x = 0 \) directly into the expression \( \frac{(e^x - 1)^2}{x \sin x} \). This yields \( \frac{(0)^2}{0 \times 0} \) or \( \frac{0}{0} \), which is an indeterminate form suitable for l'Hôpital's rule.

Step 2: Differentiate the Numerator and Denominator

Apply l'Hôpital's rule, which states that if \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) forms occur, the limit can be evaluated as \( \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \), provided the latter limit exists. Differentiate the numerator \((e^x - 1)^2 = 2(e^x - 1)e^x\). Differentiate the denominator \(x \sin x = \sin x + x \cos x\).

Step 3: Simplify the New Expression

After differentiation, the expression becomes \( \lim_{x \rightarrow 0} \frac{2(e^x - 1)e^x}{\sin x + x \cos x} \). Substitute \( x = 0 \) again to check: Numerator becomes \(2(0)e^0 = 0\) and the denominator becomes \(0 + 0 = 0\). Thus, the indeterminate form \(\frac{0}{0}\) persists, requiring another application of l'Hôpital's rule.

Step 4: Apply l'Hôpital's Rule Again

Differentiate the new numerator and denominator a second time. For the numerator: \((2(e^x - 1)e^x)' = 2(e^x(e^x - 1) + e^(2x))\). For the denominator: \((\sin x + x \cos x)' = \cos x - x \sin x + \cos x\).

Step 5: Evaluate the New Limits

The expression after the second differentiation is \( \lim_{x \to 0} \frac{2(e^x(e^x-1) + e^{2x})}{2\cos x - x\sin x} \). Substitute \( x = 0 \) into the new expression: Numerator becomes \(2(1 \times 0 + 1) = 2\); Denominator becomes \(2(1) = 2\). Thus, the limit is \( \frac{2}{2} = 1 \) after this evaluation.

Key Concepts

Indeterminate Forms

Indeterminate forms are fascinating because they appear when a straightforward calculation doesn't give us a clear answer. In calculus, the form \(\frac{0}{0}\) is a common indeterminate form. It's like a puzzle, teasing us with incompleteness when we directly substitute a value into a limit problem.

When you encounter an indeterminate form, you can't just stop there. You need other methods to proceed towards an answer. This is where techniques like L'Hôpital's Rule become extremely useful. They allow us to navigate these ambiguous situations by transforming the problem into a form we can solve.

Knowing how to identify indeterminate forms helps you recognize when a problem needs extra work. It's the catalyst for bringing L'Hôpital's Rule into play.

Differentiation

Differentiation is a process that computes the derivative of a function, which is essentially a way to measure how a function changes as its input changes. In the context of applying L'Hôpital's Rule, differentiation is crucial.

Let's consider the original function \((e^x - 1)^2 / (x \sin x)\). Direct substitution at \(x = 0\) showed an indeterminate form. To utilize L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

• The numerator, \((e^x - 1)^2\), differentiates to \(2(e^x - 1)e^x\).
• The denominator, \(x \sin x\), differentiates to \(\sin x + x \cos x\).

These transformations are vital because they convert the puzzle into a form we can unravel by evaluating a new limit.

Through differentiation, we smooth out the function's behavior around the indeterminate points, allowing us to proceed confidently to the next steps.

Limit Evaluation

Evaluating limits is a core component of calculus, and it becomes particularly engaging when dealing with indeterminate forms. In this exercise, L'Hôpital's Rule simplifies the task of limit evaluation by letting us differentiate instead of facing a dead-end.

Initially, the limit \( \lim_{x \to 0} \frac{(e^x - 1)^2}{x \sin x} \) produced \( \frac{0}{0} \), an unsolvable puzzle at first glance. By differentiating both parts a couple of times, we transformed the expression into \( \lim_{x \to 0} \frac{2(e^x(e^x-1) + e^{2x})}{2\cos x - x\sin x}\).

Substituting back \(x = 0\) into this new expression, we found concrete values for the numerator and the denominator, simplifying to \( \frac{2}{2} = 1 \). This result offers a clear answer and highlights the power of limit evaluation techniques in resolving complicated expressions.