Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0}(\cos x)^{1 / x^{2}} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{\sqrt{e}} \).

Step 1: Check if the limit is an indeterminate form

To apply l'Hôpital's Rule, we must first determine if the limit results in an indeterminate form. We'll first evaluate the base, \(\cos x\), and the exponent, \(1/x^2\), as \(x\) approaches 0. For \(x \to 0\), \(\cos x \to 1\), thereby causing `(\cos x)^{1/x^2}` to approach \(1^{\infty}\). This is indeed an indeterminate form.

Step 2: Apply the natural logarithm

To handle the indeterminate form \(1^{\infty}\), take the natural logarithm of the function: \(y = (\cos x)^{1/x^2} \Rightarrow \ln y = \frac{1}{x^2} \ln(\cos x)\). The expression \(\ln(\cos x)\) approaches \(\ln 1 = 0\) which results in the fraction \(\frac{\ln(\cos x)}{x^2}\), an indeterminate form \(\frac{0}{0}\).

Step 3: Apply l'Hôpital's Rule

Since the limit \(\lim_{x \to 0}\frac{\ln(\cos x)}{x^2}\) is of the form \(\frac{0}{0}\), we can apply l'Hôpital's Rule by differentiating the numerator and the denominator:1. The derivative of the numerator, \(\ln(\cos x)\), is \(-\tan x = -\frac{\sin x}{\cos x}\).2. The derivative of the denominator, \(x^2\), is \(2x\).This yields the new expression: \[\lim_{x \to 0}\frac{-\sin x/\cos x}{2x} = \lim_{x \to 0} -\frac{\sin x}{2x \cos x}. \]

Step 4: Evaluate the new limit

Evaluate the limit, using the well-known fact \(\lim_{x \to 0}\frac{\sin x}{x} = 1\):The new limit simplifies:\[\lim_{x \to 0} -\frac{\sin x}{2x \cos x} = -\frac{1}{2} \cdot \frac{1}{1} = -\frac{1}{2}.\]Therefore, \(\ln y\) approaches \(-\frac{1}{2}\) as \(x\) approaches 0.

Step 5: Exponentiate to find the original limit

Since \(\ln y = -\frac{1}{2}\), we return to \(y\) by exponentiating both sides:\[ y = e^{\ln y} = e^{-1/2}. \]Thus, the original limit is: \[ \lim_{x \to 0}(\cos x)^{1/x^2} = e^{-1/2} = \frac{1}{\sqrt{e}}. \]

Key Concepts

Indeterminate Forms

Indeterminate forms are situations in calculus where we cannot directly determine the limit of a function due to the form of its expression. A key example of this occurs with the expression \(1^\infty\). Here, the base approaches \(1\) while the exponent tends towards infinity. This can create ambiguity because although any number raised to the power of zero equals 1, any number raised to infinity grows boundlessly.
This results in an undefined state which needs further analysis to determine the limit.

In many cases, like the expression \((\cos x)^{1/x^2}\) as \(x\) approaches zero, the expression results in \(1^\infty\). Here, \(\cos x \to 1\) when \(x \to 0\), while \(1/x^2\) becomes infinitely large, leading to this indeterminate form.

• Recognize forms such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0\times\infty\), among others, as indeterminate.
• These forms require additional techniques such as algebraic manipulation or L'Hôpital's Rule to resolve.

Natural Logarithm

The natural logarithm, denoted \(\ln\), provides a powerful tool in calculus for dealing with complex expressions. For indeterminate forms like \(1^\infty\), taking the natural logarithm can simplify the expression by transforming powers into products.
For example, with \(y = (\cos x)^{1/x^2}\), we apply the natural logarithm to both sides, giving us \(\ln y = \frac{1}{x^2} \ln(\cos x)\). This helps convert the indeterminate exponential form into a more manageable expression that can lead to a solution.

• It converts functions of the form \(a^b\) into \(b\ln a\), simplifying differentiation and integration.
• Taking logs can simplify multiplicative relationships into additive ones, which are easier to handle.

This operation is especially useful when combined with other methods, such as L'Hôpital's Rule, to further solve and understand limits.

Exponentiation

Exponentiation is the inverse operation of taking a logarithm. It allows us to return from the logarithmic transformation back to the standard form. Once we simplify the logarithmic form using calculus techniques like L'Hôpital's Rule, exponentiation recovers the original expression.
After taking the natural logarithm and simplifying \(\ln y = -1/2\), we obtain the original limit result \(y = e^{-1/2}\) by exponentiating.

• Exponentiation solves for the original expression based on logarithmic calculations.
• It transforms solved logarithmic results back to the exponential form to find definitive solutions.

This final step is crucial as it gives us the actual limit, which, in our exercise, leads us to the conclusion that \(\lim_{x \to 0}(\cos x)^{1/x^2} = \frac{1}{\sqrt{e}}\).