Problem 36 - Essential Calculus

Question

In Exercises \(27-38,\) (a) describe the type of indeterminate form (if any) that is obtained by direct substitution. (b) Evaluate the limit, using L'Hôpital's Rule if necessary. (c) Use a graphing utility to graph the function and verify the result in part (b). \(\lim _{x \rightarrow 0^{+}}\left[\cos \left(\frac{\pi}{2}-x\right)\right]^{x}\)

Step-by-Step Solution

Answer

\(\lim _{x \rightarrow 0^{+}}\[\cos (\frac{\pi}{2}-x)\]^{x} = \infty\)

Step 1: Identify the indeterminate form

By substituting \(x = 0\) into the expression, we obtain \(\cos(\pi/2)^0\), which is a \(0^0\) indeterminate form.

Step 2: Logarithmic transformation

Transform the expression to turn it to \(0/0\) or \(\infty/\infty\) indeterminate form that is suitable to apply L'Hôpital's Rule. Let \(L = \lim _{x \rightarrow 0^{+}}\[\cos (\frac{\pi}{2}-x)\]^{x}\). Taking the natural logarithm, \(ln(L) = \lim _{x \rightarrow 0^{+}} x\ln(\[\cos (\frac{\pi}{2}-x)\])\). Now we have a new indeterminate form that is \(0 \cdot \infty\). Rewrite it to the form \(\infty/\infty\) or \(0/0\).

Step 3: Rewrite the expression

We can transform the product of \(x\) and \( \ln(\cos (\pi/2 -x))\) into a ratio. Therefore, the limit becomes: \(ln(L) = \lim _{x \rightarrow 0^{+}} \frac{ln(\[\cos (\frac{\pi}{2}-x)\])}{\frac{1}{x}}\). Now it is an indeterminate form \( \infty/ \infty\), suitable for L'Hôpital's Rule application.

Step 4: Apply L'Hôpital's Rule

By L'Hôpital's Rule, if limit has a form \(0/0\) or \(\infty/\infty\) then \(\lim_{x \to a} f(x)/g(x) = \lim_{x \to a} f'(x)/g'(x)\), \( \implies ln(L) = \lim _{x \rightarrow 0^{+}} \frac{-\sin(\frac{\pi}{2}-x)}{-\cos(\frac{\pi}{2}-x)} = \lim _{x \rightarrow 0^{+}} tan(\frac{\pi}{2}-x)\).

Step 5: Solve the transformed limit

Substitute \(x = 0\) to get \(tan(\pi/2) = \infty\). Hence, \(ln(L) = \infty\).

Step 6: Retransform and solve

Since \(ln(L) = \infty\), we can exponentiate both sides (since the exponential and the natural log are inverse functions) to get: \(L = e^{\infty} = \infty\)

Key Concepts

Indeterminate Forms

When evaluating limits, sometimes we encounter expressions that do not immediately simplify to a fixed value. These are known as indeterminate forms. An indeterminate form arises when a limit results in an expression that can’t be directly evaluated using basic algebra. Common examples include \(0/0\), \(rac{0}{0}\), and \(0^0\). These forms require special techniques to resolve, as they represent uncertainty in the mathematical expression.

In our exercise, substituting \(x = 0\) into \( ext{cos} \left( rac{\pi}{2} - x \right) )^x\) gives the \(0^0\) form. This is a classic indeterminate form because we can't simply calculate it like ordinary numbers.

To deal with such forms, we apply techniques like logarithmic transformation or L'Hôpital's Rule to make progress towards a solution.

Limits Evaluation

Evaluating limits is about finding out what a function approaches as the variable gets closer to a specific value. It's a fundamental concept in calculus and is crucial for understanding the behavior of functions at particular points.

In cases with indeterminate forms like \(0/0\) or \(rac{ ext{something}}{ ext{something}}\), L'Hôpital's Rule becomes very handy. The rule states if you have forms like these, you can take the derivative of the numerator and denominator to find the limit. In simpler terms, you differentiate the top and bottom separately and then re-evaluate the limit.

For our problem, transforming the original expression using natural logs helped shift the form from \(0^0\) to something workable like \(0/0\). With L'Hôpital's Rule, you can systematically simplify and find the limit.

Logarithmic Transformation

Logarithmic transformation is a powerful technique for making complicated problems more manageable. By applying a logarithmic transformation, we can handle exponential expressions and bring them into a form more suitable for calculus tools like L'Hôpital's Rule.

In the given problem, the expression \(\text{cos} \left( rac{\pi}{2} - x \right) )^x\) is transformed using a natural logarithm, which simplifies multiplication into addition, a more tractable form. This turns our \(0^0\) problem into a \(0 \cdot \infty\) form, which we further adjust to a ratio so that L'Hôpital's Rule applies.

After transformation, we differentiate and find the limit of \( an\left( \frac{\pi}{2} - x \right)\) as \(x\) approaches zero, which gives us a clearer path to solving the indeterminate form.