Problem 15 In Problems $1-20$, find the n... [FREE SOLUTION]

Chapter 2: Problem 15

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. $$ f(x)=\left\\{\begin{array}{cc} (\tan x) / x & x \neq 0 \\ c & x=0 \end{array}\right. $$

Short Answer

Expert verified

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

Step by step solution

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.

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 $c = 1$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Short Answer

Step by step solution

Check Continuity at x=0

Verify Differentiability at x=0

Key Concepts

Limits

Taylor Series

Trigonometric Functions

Function Continuity

Differentiable Functions

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