Problem 26 - Thomas Calculus

Question

Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{x \rightarrow(\pi / 2)}\left(\frac{\pi}{2}-x\right) \tan x $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Indeterminate Form

The given limit is \( \lim _{x \rightarrow(\pi / 2)}\left(\frac{\pi}{2}-x\right) \tan x \). As \( x \) approaches \( \pi/2 \), \( (\pi/2 - x) \) approaches 0, and \( \tan x \) approaches infinity, leading to the indeterminate form \( 0 \times \infty \). We can rewrite this as a fraction for easier handling with l'Hopital's rule.

Step 2: Rewrite as a Quotient

Rewrite the expression \( \left(\frac{\pi}{2}-x\right) \tan x \) as a fraction: \( \frac{\pi/2 - x}{1/\tan x} \). This can also be expressed as \( \frac{\pi/2 - x}{ \cot x} \), which is an indeterminate form \( 0/0 \).

Step 3: Apply l'Hopital's Rule

Since the form is \( 0/0 \), we can apply l'Hopital's Rule, which involves differentiating the numerator and the denominator:\[ \lim_{x \to \pi/2} \frac{\pi/2 - x}{\cot x} = \lim_{x \to \pi/2} \frac{-1}{-\csc^2 x}. \] Simplifying this gives \( \lim_{x \to \pi/2} \sin^2 x \).

Step 4: Evaluate the Limit

Evaluate \( \lim_{x \to \pi/2} \sin^2 x \). As \( x \) approaches \( \pi/2 \), \( \sin x \) approaches 1, so \( \sin^2 x \) approaches 1 as well. Therefore, the limit is 1.

Key Concepts

Indeterminate Forms

Indeterminate forms often occur when evaluating limits, particularly when dealing with functions tending towards zero or infinity. These forms indicate that a simple substitution into the limit expression would not yield a meaningful result. In the limit problem given, as \( x \) approaches \( \pi/2 \), the expression \( (\pi/2 - x) \tan x \) results in the form \( 0 \times \infty \). This form is "indeterminate" because multiplying zero by infinity does not provide a clear outcome. The expression must be manipulated, often by rewriting it into a fraction, to apply rules like l'Hopital's.

Common indeterminate forms include:

• \( 0/0 \)
• \( \infty/\infty \)
• \( 1^\infty \)
• \( 0^0 \)
• \( \infty - \infty \)
• \( 0 \times \infty \)

Each form signals a need for further manipulation to find a definitive limit value.

Limits

Understanding limits is crucial in calculus as they describe the behavior of functions as they approach a particular point. In evaluating \( \lim_{x \rightarrow \pi/2} \), we wanted to discover how the function \( (\pi/2 - x) \tan x \) behaved as \( x \) got very close to \( \pi/2 \). Limits help determine the "end behavior" of functions near points of interest.

When direct substitution in a limit results in an indeterminate form, as seen here, other techniques or analytical manipulations are required. This is where l'Hopital's Rule becomes handy, applied on functions rewritten as fractions in indeterminate forms like \( 0/0 \). By differentiating the numerator and the denominator separately, we attempt to simplify and directly compute the limit, leading us to a clear numeric result: here, it simplified the expression to \( \sin^2 x \), a clearer pathway to evaluate directly.

Trigonometric Functions

Trigonometric functions, such as \( \tan x \) and \( \sin x \), play a critical role in many limit problems. As \( x \) approaches \( \pi/2 \), \( \tan x \) tends to infinity, a key factor in why the original limit statement became indeterminate.

In our problem, transforming \( \tan x \) into \( \cot x \) presented the expression as \( \frac{\pi/2 - x}{\cot x} \), allowing for the application of l'Hopital's Rule. Recognizing the behavior of these trigonometric functions near specific angles is vital.

For instance, \( \sin x \) and \( \cos x \) have known values as they approach noted angles like \( \pi/2 \) or \( \pi \), which provided a decisive step in reworking the problem, resulting eventually in \( \sin^2 x \). Accurately evaluating \( \lim_{x \rightarrow \pi/2} \sin^2 x \) as \( x \) nears \( \pi/2 \) gave us 1, an essential resolution to the indeterminate form.