Problem 37 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}}(\cot x-\csc x) $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Indeterminate Form

Evaluate the individual limits of \( \cot x \) and \( \csc x \) as \( x \to 0^+ \). We know \( \cot x = \frac{\cos x}{\sin x} \) and \( \csc x = \frac{1}{\sin x} \). As \( x \rightarrow 0^+ \), both \( \frac{\cos x}{\sin x} \) and \( \frac{1}{\sin x} \) approach infinity, making this a \( \infty - \infty \) indeterminate form.

Step 2: Combine into a Single Fraction

Write \( \cot x - \csc x \) as a single fraction: \( \lim _{x \rightarrow 0^{+}} \left( \frac{\cos x}{\sin x} - \frac{1}{\sin x} \right) = \lim _{x \rightarrow 0^{+}} \frac{\cos x - 1}{\sin x} \). The expression is now in the form \( \frac{0}{0} \) as \( x \rightarrow 0^+ \).

Step 3: Apply L'Hospital's Rule

Apply L'Hospital's Rule which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) when the original limit is an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Differentiate the numerator and the denominator: \( f'(x) = -\sin x \) and \( g'(x) = \cos x \).

Step 4: Evaluate the New Limit

Evaluate the limit of the new fraction: \( \lim _{x \rightarrow 0^{+}} \frac{-\sin x}{\cos x} = \lim _{x \rightarrow 0^{+}} -\tan x \). As \( x \rightarrow 0^{+} \), \( \tan x \rightarrow 0 \). Thus, \( \lim _{x \rightarrow 0^{+}} -\tan x = 0 \).

Key Concepts

Indeterminate Forms

In calculus, understanding indeterminate forms is crucial. These forms occur when evaluating a limit produces an unconventional outcome such as \(0/0\) or \(\infty - \infty\).
In the original exercise, as \(x \to 0^+\), both \(\cot x\) and \(\csc x\) approach infinity. This results in an \(\infty - \infty\) form, which is indeterminate.
Indeterminate forms suggest that typical algebraic or substitution techniques won't work directly. Such cases often require special techniques like L'Hospital's Rule to resolve them.

• For example, the expression \(\frac{0}{0}\) signifies no value can be directly assigned without further manipulation.
• Other common indeterminate forms include \(0 \cdot \infty\), \(\infty/\infty\), and \(0^0\).

Recognizing these forms allows us to utilize specific strategies to evaluate limits.

Trigonometric Limits

Trigonometric limits specifically involve the behavior of trigonometric functions as variables approach certain values. They are essential for deeper calculus applications.
In the given problem, we deal with \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\). Both of these trigonometric functions resemble limits as \(x\) approaches 0 from the positive direction.
Trigonometric limits often lead to indeterminate forms due to the properties of sine and cosine near the origin.

• \(\sin(x)\) approaches zero as \(x\) approaches zero, which can cause denominators to lead to undefined expressions.
• Similarly, \(\cos(x)\) approaches 1 as \(x\) approaches zero, influencing limit solutions.

Careful handling of these functions with strategic techniques can resolve these limits effectively.

Calculus Techniques

In calculus, finding limits involving complex functions often requires advanced methods. One such method is L'Hospital's Rule, specifically useful for dealing with indeterminate forms.
L'Hospital's Rule helps us solve the original exercise after we construct \(\cot x - \csc x\) into a single fraction yielding \(\frac{0}{0}\) form.
This rule states that if \(f(x)\) and \(g(x)\) are differentiable and \(\lim_{x \to a} \frac{f(x)}{g(x)}\) yields an indeterminate form, then:\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]In the problem:

• The numerator \(f(x) = \cos x - 1\) has a derivative of \(-\sin x\).
• The denominator \(g(x) = \sin x\) has a derivative of \(\cos x\).

Applying these derivatives transforms the original limit into \(\lim_{x \to 0^+} \frac{-\sin x}{\cos x}\), eventually simplifying to \(0\).
Using such calculus techniques ensures a deeper understanding and simplifies the process of solving complex limit problems.