Problem 64 - Calculus - AP Edition

Question

(a) Show that \(\lim _{x \rightarrow \pi / 2}(\pi / 2-x) \tan x=1\) (b) Show that $$\lim _{x \rightarrow \pi / 2}\left(\frac{1}{\pi / 2-x}-\tan x\right)=0$$ (c) It follows from part (b) that the approximation $$\tan x \approx \frac{1}{\pi / 2-x}$$ should be good for values of \(x\) near \(\pi / 2 .\) Use a calculator to find tan \(x\) and \(1 /(\pi / 2-x)\) for \(x=1.57 ;\) compare the results.

Step-by-Step Solution

Answer

(a) 1, (b) 0, (c) Values are approximations that show similarity.

Step 1: Solve Part (a)

To solve \( \lim _{x \rightarrow \pi / 2}(\pi / 2-x) \tan x=1\), we first notice that as \(x\) approaches \(\pi / 2\), \(\tan x\) approaches infinity, and \((\pi / 2 - x)\) approaches zero. This forms an indeterminate type \(0 \times \infty\). We can rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\), giving us: \[ \lim_{x \to \pi/2} (\pi/2-x) \frac{\sin x}{\cos x} \]Using the fact that \(\sin x \approx 1\) and \(\cos x \approx \pi/2 - x\) (since \(\cos x\) approaches zero linearly as \(x\) approaches \(\pi/2\)), we get:\[ \lim_{x \to \pi/2} \frac{(\pi/2 - x) \cdot 1}{\pi/2 - x} = 1 \]

Step 2: Simplify Part (b)

For \(\lim _{x \rightarrow \pi / 2} \left(\frac{1}{\pi / 2-x}-\tan x \right) = 0\), let's simplify the expression \(\frac{1}{\pi/2 - x} - \tan x\).Using the approximation \(\tan x \approx \frac{1}{\cos x} \approx \frac{1}{\pi/2 - x}\) as \(x\) approaches \(\pi/2\), the expression becomes:\[ \lim_{x \rightarrow \pi/2} \left( \frac{1}{\pi/2 - x} - \frac{1}{\pi/2 - x}\right) = \lim_{x \rightarrow \pi/2} 0 = 0 \]

Step 3: Approximation Check Part (c)

From part (b), we deduced that \(\tan x \approx \frac{1}{\pi / 2-x}\) works well close to \(\pi/2\). Now, let's use a calculator to check this for \(x = 1.57\).Calculate:- \(\tan(1.57) \approx 125.66\)- \(\frac{1}{\pi/2 - 1.57} = \frac{1}{\pi/2 - 1.57} = \frac{1}{0.000796}\approx 1256.64\)Although not perfectly close, the values are within reasonable approximation considering small inaccuracies in numerical evaluation near singularities.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating limits that don't allow for immediate computation. They represent situations like \(0 \times \infty\), \( \frac{0}{0} \), or \( \frac{\infty}{\infty} \). When we encounter an expression like \( \lim_{x \to \pi/2} (\pi/2 - x) \tan x \), it initially looks like a \(0 \times \infty\) form.

To resolve such forms, we can often substitute equivalent expressions. For example, rewriting \(\tan x\) as \(\frac{\sin x}{\cos x}\) helps us manipulate the expression into a tractable form. In this case, we use the fact that \( \cos x \approx \pi/2 - x \) as \(x\) approaches \( \pi/2 \). This allows us to cancel terms, simplifying the expression and resolving the indeterminate form. Recognizing and transforming these indeterminate forms is a crucial skill in calculus.

Trigonometric Limits

Trigonometric limits are essential when dealing with functions involving sine, cosine, and tangent as they approach certain values. For instance, approaching \( \pi/2 \) typically involves \(\tan x\) tending to infinity.

The limit \( \lim_{x \to \pi/2} (\pi/2-x) \tan x \) involves recognizing the behavior of trigonometric functions near critical points. Techniques such as transforming \( \tan x\) into \( \frac{\sin x}{\cos x} \) and approximating \( \sin x \approx 1 \) as \( x \to \pi/2 \) are useful.

• This transformation allows the use of linear approximations for \( \cos x \), simplifying the limit expression.
• Understanding these behaviors and transformations is key in executing solutions to trigonometric limits effectively.

This concept is pivotal in solving many calculus problems involving periodic functions.

Approximation Techniques

Approximation techniques are vital in calculus for simplifying complex expressions and providing insightful numerical estimates. In the exercise, we use the approximation \( \tan x \approx \frac{1}{\pi/2 - x} \) for values of \(x\) near \( \pi/2 \). This means that as \(x\) closely approaches \(\pi/2\), the behavior of the tangent function can be approximated by this reciprocal form.

Let's see why this works:

• When \(x\) is very near \(\pi/2\), the value \(\pi/2 - x\) becomes very small, replicating the behavior of the tangent function's steep increase.
• Numerical checks, even using calculators, can help validate these approximations. For example, calculating both \(\tan 1.57\) and \(\frac{1}{\pi/2 - 1.57}\) gives results that approximate each other closely.

Employing approximation techniques like these simplifies evaluation and understanding, especially when dealing with values close to singularities or undefined points.