Problem 50 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}}(\cos (2 x))^{3 / x} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Recognize the Indeterminate Form

The limit problem \( \lim _{x \rightarrow 0^{+}}(\cos (2x))^{3/x} \) is given. As \( x \) approaches 0 from the positive side, \( \cos(2x) \) approaches 1, leading the expression inside the limit to take on the form \( 1^{\infty} \), which is an indeterminate form.

Step 2: Transform the Expression

To deal with the \( 1^{\infty} \) form, we use the natural logarithm transformation. Let \( y = (\cos (2x))^{3/x} \). Take the natural logarithm of both sides to get \( \ln y = \frac{3}{x} \ln (\cos(2x)) \).

Step 3: Rewrite the Limit in Terms of Logarithm

Rewrite the original limit problem as \( \lim _{x \rightarrow 0^{+}} \ln y = \lim _{x \rightarrow 0^{+}} \frac{3 \ln (\cos(2x))}{x} \). This now takes the form \( \frac{0}{0} \), making it suitable for L'Hospital's Rule.

Step 4: Apply L'Hospital's Rule

Apply L'Hospital's Rule. Differentiate the numerator and the denominator:Numerator derivative: \( \frac{d}{dx} [3 \ln (\cos(2x))] = 3 \times \frac{d}{dx} [\ln(\cos(2x))] = 3 \times \frac{-2 \sin(2x)}{\cos(2x)} \) = \( -6 \tan(2x) \).Denominator derivative: \( \frac{d}{dx}[x] = 1 \).Now the limit is \( \lim _{x \rightarrow 0^{+}} \frac{-6 \tan(2x)}{1} \).

Step 5: Evaluate the Limit

As \( x \) approaches 0 from the positive side, \( \tan(2x) \) approaches \( 2x \), which tends to 0. Thus, \( \lim _{x \rightarrow 0^{+}} -6 \tan(2x) = \lim _{x \rightarrow 0^{+}} -6 \cdot 2x = 0 \). Therefore, \( \lim _{x \rightarrow 0^{+}} \ln y = 0 \).

Step 6: Take the Exponential to Solve for \( y \)

Since \( \lim _{x \rightarrow 0^{+}} \ln y = 0 \), exponentiate to find the original limit:\( y = e^{\ln y} = e^{0} = 1 \).Therefore, \( \lim _{x \rightarrow 0^{+}} (\cos (2x))^{3/x} = 1 \).

Key Concepts

Indeterminate Form

In calculus, limits can sometimes lead to expressions that are difficult to evaluate directly. These are known as "indeterminate forms." They occur because, for specific input values, the expression appears undefined or ambiguous. In our exercise, the problem is to evaluate \( \lim _{x \rightarrow 0^{+}}(\cos (2x))^{3/x} \). As \( x \) approaches 0 from the positive side, the cosine term \( \cos(2x) \) tends to approach 1. This creates the form \( 1^{\infty} \), a classic example of an indeterminate form because it is not clear whether the expression grows large, stays constant, or behaves in another way. Thus, special techniques, like l'Hospital's Rule or alternative transformations, are needed to resolve this uncertainty and find the limit.

Natural Logarithm Transformation

To tackle indeterminate forms like \( 1^{\infty} \), one effective strategy is to use logarithmic transformations. By doing so, we can turn complex exponentiated forms into expressions that are easier to manage. In the exercise, this involves letting \( y = (\cos(2x))^{3/x} \) and then taking the natural logarithm (\( \ln \)) of both sides. This gives us \( \ln y = \frac{3}{x} \ln(\cos(2x)) \). The transformation converts the power to a product, which is much simpler to differentiate later. We can now redefine the problem in terms of \( \ln y \) instead of the original exponents. It shifts our focus from \( 1^{\infty} \) to a form that might be resolved using l'Hospital's Rule when both the numerator and denominator in the rewritten expression approach zero.

Differentiation

Differentiation is a fundamental tool in calculus that allows us to analyze the rate at which functions change. In our exercise, after transforming the problem with logarithms, we reach the expression \( \frac{3 \ln (\cos(2x))}{x} \), which presents the \( \frac{0}{0} \) form. This opens the door for applying l'Hospital's Rule, which involves differentiating the numerator and denominator separately.

• The derivative of the numerator \( 3 \ln(\cos(2x)) \) involves the chain rule. Hence, it becomes \(-6 \tan(2x)\).
• The denominator, \( x \), is straightforward and differentiates to 1.

Consequently, we evaluate \( \lim _{x \rightarrow 0^{+}} \frac{-6 \tan(2x)}{1} \). As \( x \) approaches 0, \( \tan(2x) \) tends to \( 2x \), leading the limit to become 0. Thus, thorough differentiation allows us to manage indeterminate forms and arrive at the correct solution.