Question

Finding a Limit of a Trigonometric Function In Exercises \(63-74,\) find the limit of the trigonometric function. $$ \lim _{h \rightarrow 0} \frac{(1-\cos h)^{2}}{h} $$

Short answer

The limit of the function as \(h\) approaches zero is 0.

Step-by-step solution

Recognize the Indeterminate Form

First, plug in \(h = 0\) to the function to obtain \(\frac{(1 - \cos 0)^2}{0}\), which simplifies to \(\frac{0}{0}\) and is an indeterminate form.

Using Trigonometric Identity

Before moving forward, it is crucial to recall an important Trigonometric identity: \(\cos^2 h + \sin^2 h = 1\). By manipulating the identity, you could find that \((1 - \cos h)^2 = 1 - 2\cos h + \cos^2 h\), which could be rewritten as \((1 - \cos h)^2 = \sin^2 h\). Now the original function becomes \(\lim _{h \rightarrow 0} \frac{\sin^2 h}{h}\)

Separate the Fraction

Now, separate the fraction into two parts which gives \(\lim _{h \rightarrow 0} \frac{\sin h}{h}\times \sin h\). The first part, \(\lim _{h \rightarrow 0} \frac{\sin h}{h}\), is a standard limit and is equal to 1 as \(h\) approaches zero.

Evaluate

Now plug \(h = 0\) in second part which is \(\sin h\), it gets \(\sin 0 = 0\). So the answer is \(1 \times 0 = 0\)

Key concepts

Indeterminate Forms

When evaluating limits in calculus, especially trigonometric functions, you often encounter expressions that seem undefined at first glance. These are referred to as "Indeterminate Forms." One of the most common is the \(\frac{0}{0}\) form, which appears when both the numerator and the denominator of a fraction approach zero. In the problem we are discussing, substituting \h = 0\ in \(\lim _{h \rightarrow 0} \frac{(1-\cos h)^{2}}{h}\) results in \(\frac{0}{0}\). This signals the need for further manipulation, rather than simply plugging in the value.

To resolve indeterminate forms:

• Apply algebraic simplification or factorization.
• Use trigonometric identities or L'Hôpital's rule if applicable.
• Analyze the behavior of the function as it approaches the limiting point.

Recognizing and appropriately dealing with these forms is crucial for correctly finding limits.

Trigonometric Identities

Trigonometric Identities are equations involving trigonometric functions that are true for every value of the variables involved. In calculus, these identities are invaluable, especially for simplifying complex expressions. In our exercise, a suitable identity helps simplify the limit expression.

One such identity necessary here is \(\cos^2 h + \sin^2 h = 1\), which rearranges to \(1 - \cos h = 2\sin^2(\frac{h}{2})\) by additional simplification. This step allows us to re-express \(1 - \cos h\) in terms of sine functions, as \(\sin^2 h\) using \( (1 - \cos h)^2 = \sin^2 h\). By using this transformation, the function becomes more approachable for solving limits.

• Simplify using identities to reduce complexity.
• Transform expressions to known limit forms.
• Facilitate evaluation of indeterminate forms.

Knowing and applying these identities is essential for manipulating trigonometric expressions effectively.

Calculus

Calculus is the mathematical study of continuous change, and it provides powerful tools for finding limits and handling indeterminate forms. Fundamental techniques in calculus, such as limit evaluation, are central to solving problems like the one presented.

Here are a few steps involved in using calculus to solve the given problem:

• First, recognize that direct substitution yields an indeterminate form.
• Use a known limit, such as \(\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1\), to transform complex expressions into simpler components.
• Evaluate the function as it approaches the parameter of interest, leveraging continuity and limits.

By separating the given function into manageable parts — specifically rewriting \(\lim _{h \rightarrow 0} \frac{\sin^2 h}{h}\) as \(\lim _{h \rightarrow 0} \frac{\sin h}{h} \times \sin h\) — calculus helps clarify the behavior near the limits, making it easier to resolve the original indeterminate form.