Question

(a) Prove: If \(f\) is continuous at \(x_{0}\) and there are constants \(a_{0}\) and \(a_{1}\) such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)}{x-x_{0}}=0 $$ then \(a_{0}=f\left(x_{0}\right), f^{\prime}\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=a_{1} .\) (b) Give a counterexample to the following statement: If \(f\) and \(f^{\prime}\) are continuous at \(x_{0}\) and there are constants \(a_{0}, a_{1}\). and \(a_{2}\) such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)-a_{2}\left(x-x_{0}\right)^{2}}{\left(x-x_{0}\right)^{2}}=0 $$ then \(f^{\prime \prime}\left(x_{0}\right)\) exists.

Short answer

#Answer (a) We proved that \(a_0 = f(x_0)\), the function is differentiable at \(x_0\), and \(f'(x_0) = a_1\). (b) The function \(f(x) = |x|\) with \(x_0=0\), \(a_0=0\), \(a_1=0\) and \(a_2=\frac{1}{2}\) serves as a counterexample, as the limit expression exists and equals \(0\) when \(x\) approaches \(0\), but \(f''(x)\) does not exist at \(x=0\).

Step-by-step solution

Rewrite the given limit expression

We are given the limit expression: $$ \lim_{x \rightarrow x_{0}} \frac{f(x) - a_0 - a_1(x - x_{0})}{x - x_{0}} = 0 $$ We can rewrite the expression as: $$ \lim_{x \rightarrow x_{0}} \frac{f(x) - a_0 - a_1(x - x_{0})}{x - x_{0}} - \lim_{x \rightarrow x_{0}} a_1 = 0-\lim_{x \rightarrow x_{0}} a_1 $$

Use the definition of derivative

The derivative of a function \(g(x)\) is defined as: $$ g'(x_0) = \lim_{x \rightarrow x_0} \frac{g(x) - g(x_0)}{x - x_0} $$ Given that \(\lim_{x \rightarrow x_0} \frac{f(x) - a_0 - a_1(x - x_0)}{x - x_0} - a_1 = 0\), we have, $$\lim_{x \rightarrow x_0} \frac{f(x) - a_0 - a_1(x - x_0)}{x - x_0} = a_1$$ Let \(g(x) = f(x) - a_1(x - x_0)\). Then this expression becomes: $$ \lim_{x \rightarrow x_0} \frac{g(x) - g(x_0)}{x - x_0} = a_1 $$ By the definition of derivative, this means that \(g'(x_0) = a_1\).

Prove \(a_0=f(x_0)\)

From step 2, we know that \(g'(x_0) = a_1\). This implies that \(g(x) = f(x) - a_1(x - x_0)\) is differentiable at \(x_0\). Since \(f(x)\) is continuous at \(x_0\), it follows that the function \(g(x) = f(x) - a_1(x - x_0)\) is also continuous at \(x_0\). Evaluating \(g(x_0)\), we get $$ g(x_0) = f(x_0) - a_1(x_0 - x_0) = f(x_0) $$ Since \(g(x)\) is continuous at \(x_0\), we have \(g(x_0) = a_0\). Therefore, \(a_0 = f(x_0)\).

Prove \(f'(x_0) = a_1\)

From step 2, we already showed that \(g'(x_0) = a_1\). Using the definition of \(g(x)\), we can find the derivative of \(f(x)\). We have, $$ f'(x_0) = g'(x_0) + a_1'(x - x_0) = g'(x_0) = a_1 $$ So, \(f'(x_0) = a_1\), and we have proved the desired result for part (a). #Part (b)#

Provide a counterexample

Let's consider the function $$ f(x) = |x| $$ which is continuous at all \(x\). The derivative of \(f(x)\) is: $$ f'(x) = \begin{cases} 1 & x > 0 \\ -1 & x < 0 \end{cases} $$ which is also continuous at all \(x\) except \(x=0\). Now let's choose \(x_0=0\), \(a_0=0\), \(a_1=0\) and \(a_2=\frac{1}{2}\). Then the limit expression becomes: $$ \lim _{x \rightarrow 0} \frac{|x|-0\left(x\right)-\frac{1}{2}\left(x\right)^2}{\left(x\right)^2} = \lim _{x \rightarrow 0} \frac{|x|}{x^2} $$ This limit exists and equals \(0\) when \(x\) approaches \(0\). However, \(f''(x)\) does not exist at \(x = 0\) because \(f'(x)\) has a jump discontinuity at \(x=0\). Therefore, we have provided a counterexample for part (b).

Key concepts

Limit Definition of Derivative

In real analysis, the concept of a function's derivative is rigorously defined using the limit process. The limit definition of the derivative encompasses the fundamental idea that the derivative at a point measures the rate at which the function value changes as the input approaches that point.

To understand this definition, consider a function \(f\) that is defined near a point \(x_0\). The derivative of \(f\) at \(x_0\), denoted by \(f'(x_0)\), is given by the limit:
\[ f'(x_0) = \textrm{lim}{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0} \]
If this limit exists, it represents the slope of the tangent line to the function \(f\) at \(x_0\). It's important for students to grasp that when the limit yields a finite value, the function is said to be differentiable at \(x_0\), possessing a well-defined tangent line at that point. In contrast, if the limit does not exist or is infinite, \(f\) is not differentiable at \(x_0\).

Continuity of Functions

Continuity is a fundamental concept in calculus and real analysis that describes the behavior of functions as they approach particular points. A function \(f\) is continuous at a point \(x_0\) if the following conditions are met:

• The function \(f\) is defined at \(x_0\).
• The limit of \(f(x)\) as \(x\) approaches \(x_0\) exists.
• The limit equals the function value at that point: \(\lim_{x \rightarrow x_0} f(x) = f(x_0)\).

Simply put, a function is continuous at a point if there are no breaks, jumps, or holes in the graph of the function at that point. It's crucial to understand that continuity is a prerequisite for differentiability; a function must be continuous at \(x_0\) to have a derivative there. However, the converse is not always true: a function can be continuous at a point but not differentiable there (for example, a sharp 'corner' or 'cusp' on the graph at the point).

Differentiability

Differentiability is a property of a function that indicates whether a function has a derivative at a given point. A function \(f\) is differentiable at a point \(x_0\) if the derivative \(f'(x_0)\) exists, which means that the function's graph has a non-vertical tangent line at that point.

To illustrate differentiability using the limit definition, consider the expression \(\textrm{lim}_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0}\). This expression describes the slope of the secant line between points \((x_0, f(x_0))\) and \((x, f(x))\). As \(x\) approaches \(x_0\), if the secant lines approach a unique slope, the function is differentiable and that slope is the derivative. Geometrically, this means the graph smoothens out to a single, definitive tangent line as you zoom in on \(x_0\). Notably, if a function is differentiable at \(x_0\), it is also continuous there, but the reverse may not hold true.

Counterexamples in Real Analysis

Counterexamples play a vital role in real analysis by demonstrating the boundaries of mathematical theorems. These are specific cases that show an assertion to be false, which forces the refinement of the conditions under which the assertion can hold true.

In the context of derivatives, a classic counterexample is the function \(f(x) = |x|\) at \(x = 0\). Although \(f(x)\) is continuous at \(x = 0\), its derivative is not defined at that point because the function takes a sharp turn, resulting in different left- and right-hand side limits of the difference quotient. Hence, \(f(x)\) is not differentiable at \(x = 0\). Such counterexamples are essential learning tools as they clarify the necessity of every condition in a definition or theorem. In particular, they teach that while a function may seem to have a certain property at first glance, careful analysis is required to ascertain the true nature of its behavior.