Question

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

Short answer

The limit is 1.

Step-by-step solution

Rewrite the Expression

We start by rewriting the given expression \(( ext{sin } x)^{ ext{cos } x} = e^{ ext{cos } x imes ext{ln}( ext{sin } x)}\)for easier differentiation.Therefore, we're looking for\( ext{lim}_{x\rightarrow \pi/2} ext{cos } x \cdot ext{ln}( ext{sin } x).\)

Check the Form

Substitute \(x = \pi/2\) into the expression:- \(\cos(\pi/2) = 0\)- \(\ln(\sin(\pi/2)) = \ln(1) = 0\).The form is \(0 \times 0 = 0\). To apply l'Hôpital's Rule, it should be either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Transform to Indeterminate Form

Rewrite the expression as a quotient:\[\lim_{x \rightarrow \pi/2} \frac{\ln(\sin x)}{1/\cos x}.\]Now, substitute \(x = \pi/2\) to confirm the form:- Numerator: \(\ln(\sin(\pi/2)) = 0\)- Denominator (as \(1/\cos(\pi/2)\) approaches infinity): infinite.The expression now has the indeterminate form \(\frac{0}{0}\).

Apply l'Hôpital's Rule

Since the expression is in the indeterminate form \(\frac{0}{0}\), apply l'Hôpital's Rule:\[\lim_{x \rightarrow \pi/2} \frac{\ln(\sin x)}{1/\cos x} = \lim_{x \rightarrow \pi/2} \frac{(\ln(\sin x))'}{(1/\cos x)'}\]Calculate the derivatives:- Derivative of \(\ln(\sin x)\) is \(\frac{\cos x}{\sin x}\) or \(\cot x\).- Derivative of \(1/\cos x\) (or \(\sec x\)) is \(\sec x \tan x\).

Simplify and Evaluate the Limit

Now evaluate the limit:\[\lim_{x \rightarrow \pi/2} \frac{\cot x}{\sec x \tan x} = \lim_{x \rightarrow \pi/2} \frac{\frac{\cos x}{\sin x}}{(\sec x)(\tan x)}\]Rewriting the tan and sec terms:- \(\sec x = \frac{1}{\cos x}\)- \(\tan x = \frac{\sin x}{\cos x}\)This simplifies to:\[\lim_{x \rightarrow \pi/2} \frac{\cot x}{\sec x \tan x} = \lim_{x \rightarrow \pi/2} \frac{\cos^2 x}{\sin^2 x}\]As \(x\) approaches \(\pi/2\), \(\cos x \to 0\) and \(\sin x \to 1\):\[\lim_{x \rightarrow \pi/2} \frac{0^2}{1^2} = 0\]Hence, the limit of the original expression is:\(e^0 = 1\).

Conclusion

Thus, the limit of \((\sin x)^{\cos x}\) as \(x\) approaches \(\pi/2\) is \(1\). The entire process involved handling the initial 0×0 form of the inner exponential expression, converting to a \(\frac{0}{0}\) form, and applying l'Hôpital's rule appropriately.

Key concepts

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating limits leads to ambiguous expressions that do not directly signal a specific value. These forms typically include expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and other variations, such as \( 1^\infty \) or \( 0^0 \). Essentially, they mean that if you merely substitute the limit point into the function, it gives no clear information about what the limit is.
To solve such expressions, especially in the context of l'Hôpital's Rule, the primary goal is to transform these indeterminate forms into a form suitable for analysis. For example, in the exercise given, initially substituting \( x = \pi/2 \) resulted in a form \( 0 \times 0 \), which is not suitable for l'Hôpital's Rule. By manipulating the expression into \( \frac{0}{0} \), we make it amenable for deeper analysis.
L'Hôpital's Rule becomes applicable with these transformations, allowing us to determine limits by differentiating the numerator and denominator. Successfully navigating these transformations and manipulations is key to resolving limits in calculus.

Limits in Calculus

The concept of limits is fundamental to calculus as it describes what happens to a function as the input approaches a particular value. Limits allow us to analyze the behavior of functions without necessarily reaching the point of interest directly. This is crucial when dealing with points where a function might not be directly defined, or yields an indeterminate form.
Limits are typically solved by direct substitution, but many functions result in indeterminate forms that require special strategies. L'Hôpital's Rule is one of these strategies, offering a method to find limits with indeterminate forms like \( \frac{0}{0} \). By taking derivatives of the numerator and denominator, we transform the problem to a simpler form that can be resolved by straightforward substitution.
Being comfortable with limits and their associated rules is important in understanding function behavior, continuity, and even the foundational rules of calculus such as derivatives and integrals.

Exponential Functions

Exponential functions, characterized by their constant base and variable exponent, are essential in many real-world applications. In mathematical analysis, they exhibit unique properties like constant growth rate per interval, making them crucial in modeling scenarios involving growth and decay.
The problem poses an expression \((\sin x)^{\cos x}\), which can be complex to handle directly. By rewriting the expression as an exponential \( e^{\cos x \cdot \ln(\sin x)} \), we convert it into a form that makes differentiation straightforward. This transformation simplifies the process of finding the limit, utilizing the properties of logarithms and exponentials.
Understanding how to manipulate exponential functions and interpret their limits provides insights into more complex functions and their behavior at specific points, expanding your calculus toolkit significantly.