Question

\(\lim _{\mathrm{x} \rightarrow(\pi / 8)}(\sin 4 \mathrm{x})^{(\tan ) 2(4 \mathrm{x})}=\ldots \ldots .\) (a) \(\mathrm{e}^{(1 / 4)}\) (b) \(\mathrm{e}^{[(-1) / 2]}\) (c) \(\mathrm{e}^{[(-1) / 4]}\) (d) \(\mathrm{e}^{(1 / 2)}\)

Short answer

The short answer for the given question is: \[\lim_{x\to\frac{\pi}{8}} (\sin 4x)^{(\tan^2 4x)} = e^{-\frac{1}{2}}\]

Step-by-step solution

Analyze the given expression and identify the indeterminate form

First, let's analyze the given expression: \(\lim_{x\to \frac{\pi}{8}} (\sin 4x)^{(\tan^2 4x)}\) As \(x\) approaches \(\pi/8\), the sine function becomes: \(\sin\left(\frac{\pi}{2}\right) = 1\) Meanwhile, the tangent squared function becomes: \(\tan^2\left(\frac{\pi}{2}\right) = \infty\) Since the expression is in the form of \(1^{\infty}\), which is an indeterminate form, we will need to apply L'Hôpital's rule using natural logarithm.

Apply natural logarithm to the given expression

To prepare the expression for L'Hôpital's rule, we will first take the natural logarithm on both sides of the equation: \(y = (\sin 4x)^{(\tan^2 4x)}\) \(\ln y = (\tan^2 4x) \cdot \ln(\sin 4x)\) Now, we can apply L'Hôpital's rule to the expression \(\frac{\ln (\sin4x)}{\frac{1}{\tan^2 4x}}\) until we have an answer.

Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and the denominator of the expression with respect to x and then evaluate the limit: \(\lim_{x\to \frac{\pi}{8}} \frac{\ln (\sin4x)}{\frac{1}{\tan^2 4x}} = \lim_{x\to \frac{\pi}{8}} \frac{\frac{d}{dx} \ln (\sin4x)}{\frac{d}{dx} \frac{1}{\tan^2 4x}}\) Now, using the chain rule, we find that: \(\frac{d}{dx} \ln (\sin4x) = \frac{4\cos4x}{\sin4x}\) \(\frac{d}{dx} \frac{1}{\tan^2 4x} = -8\sec^2 4x\tan 4x\) Plugging back into the limit: \(\lim_{x\to \frac{\pi}{8}} \frac{\frac{4\cos4x}{\sin4x}}{-8\sec^2 4x\tan 4x}\) Now, we can simplify this expression and evaluate the limit: \(\lim_{x\to \frac{\pi}{8}} \frac{\frac{4\cos4x}{\sin4x}}{-8\left(\frac{1}{\cos^2 4x}\right)\left(\frac{\sin 4x}{\cos 4x}\right)} = \lim_{x\to \frac{\pi}{8}} \frac{-1}{2}\) So, the limit of the expression after applying natural logarithm is \(\frac{-1}{2}\).

Find the original limit

Now, remember that we took the natural logarithm of our original limit expression. To find the original limit, we must apply the inverse function, exponentiation, to the result we just found: \(e^{\lim_{x\to \frac{\pi}{8}} \left[-\frac{1}{2}\right]} = e^{-\frac{1}{2}}\) Therefore, the limit for the given exercise is: \(\lim_{x\to\frac{\pi}{8}} (\sin 4x)^{(\tan^2 4x)} = e^{-\frac{1}{2}}\) The correct answer is option (b).

Key concepts

Indeterminate Forms

When we talk about indeterminate forms in calculus, we are dealing with expressions that don’t directly allow us to determine a limit. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\). These forms often arise when substituting a value into a limit and obtaining an undefined result.

In the given exercise, as \(x\) approaches \(\pi/8\), the expression \((\sin 4x)^{(\tan^2 4x)}\) reduces to the form \(1^{\infty}\). At this point, the expression does not give us a straightforward answer. That's where applying techniques like L'Hôpital's Rule can help us resolve these indeterminate forms and find a meaningful limit.

Resolving indeterminate forms often involves algebraic manipulation, such as factoring or using logarithms, to transform the expression into a form conducive to limit evaluation.

Limits in Calculus

Limits are foundational in calculus, and they help us understand the behavior of functions as they approach specific points or infinity. The notation \(\lim_{x \to a} f(x)\) describes the value that the function \(f(x)\) gets closer to as \(x\) gets nearer to \(a\).

Calculating limits can involve direct substitution, or require special techniques for cases of indeterminate forms. The direct evaluation works well for continuous functions, where plugging in \(x = a\) gives the limit immediately. However, when dealing with indeterminate forms, limits need more sophisticated methods like L'Hôpital's Rule.

In the provided problem, determining the limit of the function \((\sin 4x)^{(\tan^2 4x)}\) involves transforming the problem into one solvable through calculus techniques. This necessitates applying the natural logarithm function, making the function easier to differentiate and work with mathematically.

Natural Logarithm Application

The natural logarithm, denoted as \(\ln\), is a powerful tool in calculus, particularly useful in solving limits involving exponentiation. By converting an exponential expression to a logarithmic one, we can ease the process of applying calculus techniques like differentiation.

In the exercise, the expression \((\sin 4x)^{(\tan^2 4x)}\) is transformed by taking the natural logarithm on both sides to get \(\ln y = (\tan^2 4x) \cdot \ln(\sin 4x)\). This transformation allows us to apply L'Hôpital's Rule, as it converts the expression into a quotient form. The logarithm helps by linearizing the exponential expression, making it easier to handle analytically.

After applying L'Hôpital's Rule and calculating the limit in the logarithmic domain, the answer requires exponentiating the result to get back to the original expression. Consequently, this gives us the value of \(e^{-\frac{1}{2}}\), demonstrating the effectiveness of using the natural logarithm in calculus problems involving limits.