Problem 47 - Thomas Calculus

Question

Is there a value of \(c\) that will make $$ f(x)=\left\\{\begin{array}{cc}{\frac{\sin ^{2} 3 x}{x^{2}},} & {x \neq 0} \\\ {c,} & {x=0}\end{array}\right. $$ continuous at \(x=0 ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, \(c = 9\) makes the function continuous at \(x = 0\).

Step 1: Understand the Problem

To determine if the function \(f(x)\) is continuous at \(x = 0\), we need to check if the limit of \(f(x)\) as \(x\) approaches 0 from both directions equals \(f(0) = c\). This means finding \(\lim_{x \to 0} \frac{\sin^2(3x)}{x^2}\).

Step 2: Use Trigonometric Limit Identity

Recall the identity \(\sin(x)/x \to 1\) as \(x \to 0\). Rewrite \(\frac{\sin^2(3x)}{x^2}\) as \(\left(\frac{\sin(3x)}{x}\right)^2\). To use the identity, we have: \(\frac{\sin(3x)}{x} = \frac{\sin(3x)}{3x} \cdot 3\).

Step 3: Apply Limit to Simplify

Appy the limit: \(\lim_{x \to 0} \left(\frac{\sin(3x)}{3x}\right)^2 \cdot 3^2 = \lim_{x \to 0} (1)^2 \cdot 9 = 9\). Thus, the limit of \(\frac{\sin^2(3x)}{x^2}\) as \(x\) approaches 0 is 9.

Step 4: Compare Limit with Function Value

For the function to be continuous at \(x = 0\), the limit we found (9) must equal \(f(0) = c\). Therefore, set \(c = 9\) to make the function continuous at \(x = 0\).

Key Concepts

Trigonometric Limits

Trigonometric limits are crucial in calculus, especially when dealing with the continuity of functions involving trigonometric expressions. One important identity to remember is:

• For a small angle x, the limit \(\sin(x)/x \to 1\) as \(x \to 0\).

This identity is very helpful when simplifying complex expressions that involve trigonometric functions.
In the given exercise, we have the expression \(\frac{\sin^2(3x)}{x^2}\), which can be difficult to evaluate directly as \(x\) approaches zero. By using the trigonometric identity, we rewrite the expression in a form that we can easily manage: \(\left(\frac{\sin(3x)}{3x}\right)^2\).
This simplification allows us to replace the fraction \(\frac{\sin(3x)}{3x}\) with 1 as \(x\to 0\), simplifying the process of limit evaluation.

Piecewise Functions

Piecewise functions are functions composed of different expressions, each valid for a specific part of the domain. They are useful for modeling situations where an expression changes based on the input value.
In our exercise, the function is defined as:

• \(f(x) = \frac{\sin^2(3x)}{x^2}\), when \(x eq 0\)
• \(f(x) = c\), when \(x = 0\)

To ensure the entire function is continuous at the transition point (in this case, \(x=0\)), the limit of the expression as \(x\) approaches this point must equal the function's defined value at that point (\(c\)).
The use of piecewise functions allows us to set specific values at points which might otherwise be undefined or problematic to evaluate.

Limit Evaluation

Evaluating limits is a fundamental part of calculus, helping determine the behavior of functions as the input approaches certain values. To ensure continuity at a point, we must check that the limit of the function as it approaches from the left matches the limit as it comes from the right.
For the current function, \(f(x)=\frac{\sin^2(3x)}{x^2}\) when \(xeq 0\), the task was to determine \(\lim_{x \to 0} f(x)\).
A clever strategy for evaluating complex limits is to simplify the expression using known mathematical identities. Here, utilizing \(\frac{\sin(3x)}{3x} \to 1\), we are able to simplify the problematic term and find the limit: 9.

• We were able to split \(\lim_{x \to 0} \left(\frac{\sin(3x)}{3x}\right)^2 \cdot 9\) into parts we could evaluate separately.
• Once simplified, the limit yields a single value (9), indicating that for continuity at \(x=0\), the function's value here must also be 9.