Problem 15 - Calculus

Question

In Problems \(1-20\), find the numbers \(c\) that make \(f(x)\) into (A) a continuous function and (B) a differentiable function. In one case \(f(x) \rightarrow f(a)\) at every point, in the other case \(\Delta f / \Delta x\) has a limit at every point. $$ f(x)=\left\\{\begin{array}{cc} (\tan x) / x & x \neq 0 \\ c & x=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The number \(c = 1\) makes \(f(x)\) continuous and differentiable.

Step 1: Check Continuity at x=0

For continuity at a point, the function must be defined at that point, and the left-hand and right-hand limits at that point must equal the function's value. Thus, for continuity at \(x = 0\), we need \(\lim_{x \to 0} f(x) = f(0) = c\). We first check \(\lim_{x \to 0} (\tan x)/x\). Using the limit property, \(\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x \cos x} = \lim_{x \to 0} \frac{1}{\cos x} \cdot \lim_{x \to 0} \frac{\sin x}{x} = 1 \), since \(\frac{\sin x}{x}\to 1 \) and \(\cos 0 = 1\). Therefore, \(c = 1\) for \(f(x)\) to be continuous.

Step 2: Verify Differentiability at x=0

For differentiability, \(f(x)\) must first be continuous at \(x=0\), which we have ensured by setting \(c=1\). Now, we need \(\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}\) to exist, which is \(\lim_{x \to 0} \frac{(\tan x)/x - 1}{x}\). Simplifying, consider \(g(x) = \frac{\tan x}{x}\) and evaluate the limit \(\lim_{x \to 0} \frac{g(x)-1}{x}\). Using Taylor expansion, \(\tan x \approx x + \frac{x^3}{3}\) near zero gives \(\frac{\tan x}{x} \approx 1 + \frac{x^2}{3}\). Thus, \(\lim_{x \to 0} \frac{g(x)-1}{x} = \lim_{x \to 0} \frac{x^2/3}{x} = 0\), so the function \(f(x)\) is differentiable at \(x = 0\) when \(c = 1\).

Key Concepts

Limits

Understanding limits is crucial when discussing continuity and differentiability. A limit helps identify the behavior of a function as the input approaches a particular point. Consider a function \( f(x) \). To find if \( f(x) \) is continuous at a point \( x = a \), we need \( \lim_{x \to a} f(x) \) to equal \( f(a) \).
In our exercise, for \( f(x) = \frac{\tan x}{x} \), we are concerned with the limit as \( x \to 0 \). We evaluate \( \lim_{x \to 0} \frac{\tan x}{x} \) using known limit properties.

• \( \tan x \approx x + \frac{x^3}{3} + \ldots \) for small \( x \).
• This approximation shows that \( \frac{\tan x}{x} \to 1 \) as \( x \to 0 \).

This limit verification ensures that \( c = 1 \) makes \( f(x) \) continuous at zero.

Taylor Series

The Taylor series expansion gives us a powerful way to approximate complex functions near a specific point. For small \( x \), functions like trigonometric and exponential can be expanded into a series to simplify limits and derivatives.
For \( \tan x \), the Taylor expansion around zero is \( x + \frac{x^3}{3} + \mathcal{O}(x^5) \).
This expands \( \frac{\tan x}{x} \approx 1 + \frac{x^2}{3} \), showing how close \( \tan x \) is to \( x \) at small values.
Using Taylor series, we can simplify the expression and effectively find limits like in our exercise where we checked

• \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)
• Ensures the function's smooth behavior in close proximity to \( x = 0 \).

Thus, Taylor series helps us in proving continuity at \( x = 0 \) effectively.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are fundamental in calculus due to their periodic nature and rules.
The function \( \tan x \) is particularly special because it combines properties from sine and cosine:
\( \tan x = \frac{\sin x}{\cos x} \). As students, it's important to remember these basic trigonometric identities which help simplify problems involving derivatives and limits.
In calculus, when dealing with limits, the small angle approximations for \( \sin x \approx x \) and \( \cos x\approx 1 \) are crucial:

• They aid in quickly identifying the behavior of \( \tan x \) for values close to zero.
• Facilitate calculations by reducing complexity, as shown when evaluating \( \frac{\tan x}{x} \).
  

This understanding is central to solving our exercise efficiently.

Function Continuity

Continuity is a fundamental property for functions, ensuring a function's output behaves predictably as inputs vary smoothly.
A continuous function doesn't jump or break at a given point, \( a \), which mathematically means:

• The function's value \( f(a) \) is defined.
• \( \lim_{x \to a} f(x) = f(a) \).

For a function to be continuous, these conditions must hold at every conceivable point within its domain.
In our exercise, for \( x = 0 \), ensuring that \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \) means assigning \( c = 1 \) satisfies the continuity definition:

• This equality ensures no abrupt change in behavior near zero.
• Our approach always considers left-hand and right-hand limits to verify this aspect.

Understanding these fundamentals helps in identifying where and how a function maintains constant behavior.

Differentiable Functions

A function, which is not only continuous but also has a well-defined tangent (derivative) at a point, is called differentiable at that point. For differentiability, the function must be smooth, with no sharp corners or cusps.
Mathematically, \( f(x) \) is differentiable at \( x = a \) if \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) exists.
The differentiability implies continuity but not vice-versa.

• In our exercise, after confirming continuity at \( x = 0 \),
• We check \( \lim_{x \to 0} \frac{{\frac{\tan x}{x} - 1}}{x} = 0 \) indicating differentiability.

This step confirms \( c = 1 \) leads to both continuous and differentiable behavior, forming a harmonious, complete function at \( x = 0 \).
Integrating these concepts is crucial for advanced calculus problems where differentiability gives more detailed insights into function behavior.