Problem 47 - Thomas Calculus

Question

Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)^{2}}{x \sin x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Check the Indeterminate Form

Evaluate the limit by substituting 0 directly:- The numerator becomes \((e^0 - 1)^2 = 0^2 = 0\).- The denominator is \(x \sin x\) which also evaluates to \(0 \times \sin 0 = 0\).Since the form is \( \frac{0}{0} \), we can use l'Hopital's rule.

Step 2: Apply l'Hopital's Rule

l'Hopital's rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) is an indeterminate form \( \frac{0}{0} \), then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \] So, take the derivatives of the numerator and the denominator.

Step 3: Find the Derivatives

Calculate the derivative of the numerator and denominator:- For \(f(x) = (e^x - 1)^2\), use the chain rule: \[ f'(x) = 2(e^x - 1) \cdot e^x \]- For \(g(x) = x \sin x\), use the product rule: \[ g'(x) = x \cos x + \sin x \]

Step 4: Evaluate the Limit of Derivatives

Substitute the derivatives back into the limit:\[ \lim_{x \to 0} \frac{2(e^x - 1)e^x}{x \cos x + \sin x} \]Plug \(x = 0\) into the modified expression:- In the numerator: \(2(e^0 - 1)e^0 = 2(0) = 0\)- In the denominator: \(x \cos x + \sin x\) simplifies to \(0 \cdot 1 + 0 = 0\)As the form remains \(\frac{0}{0}\), apply l'Hopital's rule again.

Step 5: Repeat l'Hopital's Rule

Differentiate the numerator and the denominator once more:- The derivative of \( f'(x) = 2(e^x-1)e^x \) using the chain and product rules is: \[ f''(x) = 2[e^x \cdot e^x + (e^x - 1)e^x] = 4e^x(e^x - 0.5) \] - The derivative of \( g'(x) = x \cos x + \sin x \) using the product rule is: \[ g''(x) = \cos x - x \sin x + \cos x = 2 \cos x - x \sin x \]Substitute back into the limit:\[ \lim_{x \to 0} \frac{4e^x(e^x - 0.5)}{2 \cos x - x \sin x} \]

Step 6: Evaluate the Simplified Limit

Now, evaluate the limit as \(x\) approaches 0:- In the numerator, \(4e^0(e^0 - 0.5) = 4(1)(1 - 0.5) = 4 \times 0.5 = 2\)- In the denominator, \(2 \cos 0 - 0 \cdot \sin 0 = 2 \cdot 1 = 2\)The limit simplifies to \( \frac{2}{2} = 1 \).

Key Concepts

Indeterminate Forms

When dealing with limits, certain expressions can be tricky, especially when you encounter what are known as indeterminate forms. These forms are not straightforward because they don't have a clear value or behavior as they stand. Common indeterminate forms include \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).

In the problem we're exploring, when we plugged in \(x = 0\), the expression \( (e^x - 1)^2 / (x \sin x) \) resulted in both the numerator and denominator equating to zero. This situation is the classic \( \frac{0}{0} \) indeterminate form. To solve such limits, we need to apply techniques that simplify the expression, allowing us to compute the limit accurately. This is where L'Hôpital's Rule comes into play, which allows us to find the limit by differentiating the numerator and the denominator.

Differentiation

Differentiation is a method used in calculus to find how a function changes as its input changes. It's all about calculating the rate of change or slope of the function at any given point. Differentiation is a core concept when applying L'Hôpital's Rule, because to use the rule, we need to differentiate the numerator and the denominator of the limit expression separately.

In our exercise, we differentiated the numerator \((e^x - 1)^2\) using the chain rule. The chain rule helps us differentiate composite functions by dealing with each layer of the function separately. We applied the rule to get the derivative \( f'(x) = 2(e^x - 1)e^x \).

For the denominator \(x \sin x\), we used the product rule. The product rule is useful when differentiating products of two functions, and it gave us \( g'(x) = x \cos x + \sin x \). Differentiating accurately is crucial for L'Hôpital's Rule to work as intended.

Limits

The concept of limits is foundational to calculus and helps us understand the behavior of a function as it approaches a certain point. A limit examines what value a function approaches as the input gets closer to a specific number.

In solving our exercise, we sought to evaluate \( \lim_{x \rightarrow 0} \frac{(e^x-1)^2}{x \sin x} \) using L'Hôpital’s Rule. Initially, direct substitution led to an indeterminate \( \frac{0}{0} \) form, indicating more work was needed to find the true limit.

Through differentiation and applying L'Hôpital’s Rule, we increased the clarity of the limit’s behavior, eventually resolving it to a precise value. Understanding limits as approaching behavior—rather than points of arrival—offers deep insight into how functions behave locally.

Exponential Functions

Exponential functions are of the form \(e^x\) where \(e\) is a mathematical constant approximately equal to 2.718. These functions are key in many areas of mathematics due to their unique growth properties. In our problem, the function \(e^x\) is part of the expression \((e^x - 1)^2\) in the numerator.

Exponential functions have a neat differentiation property: the derivative of \(e^x\) is \(e^x\) itself, which simplifies calculations. This property made our differentiation process easier when applying L'Hôpital’s Rule because the exponential function is predictable and behaves smoothly across its domain.

This quality is also why exponential functions appear frequently in calculus problems—they are both versatile and reliable when modeling real-world scenarios or abstract mathematical concepts.