Problem 62 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Show that if a function \(f\) is differentiable at \(x=a\), then \(f\) must be continuous at that number. Hint: Write $$ f(x)-f(a)=\left[\frac{f(x)-f(a)}{x-a}\right](x-a) $$ Use the product rule for limits and the definition of the derivative to show that $$ \lim _{x \rightarrow a}[f(x)-f(a)]=0 $$

Step-by-Step Solution

Answer

We'll prove that a function \(f\) is continuous at \(x=a\) if it is differentiable at \(x=a\). We know that the function is differentiable at \(x=a\) if the following limit exists: $$ f'(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} $$ Rewriting the function as shown: $$ f(x)-f(a)=\left[\frac{f(x)-f(a)}{x-a}\right](x-a) $$ We want to show that: $$ \lim _{x \rightarrow a}[f(x)-f(a)] = 0 $$ Using the product rule for limits: $$ \lim _{x \rightarrow a}[f(x)-f(a)] = \lim _{x \rightarrow a}h(x) \times \lim_{x \rightarrow a}g(x) =f'(a) \times 0 = 0 $$ Since the limit of \([f(x)-f(a)]\) as \(x\) approaches \(a\) is equal to 0, we have proved that if a function \(f\) is differentiable at \(x=a\), it must also be continuous at that point.

Step 1: Rewrite the function

We will first rewrite the function \(f(x)\) in the given form: $$ f(x)-f(a)=\left[\frac{f(x)-f(a)}{x-a}\right](x-a) $$

Step 2: Definition of derivative

At the point \(x=a\), the function \(f\) has a derivative \(f'(a)\) if the following limit exists: $$ f'(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} $$

Step 3: Product rule for limits

We have the two functions from the given equation: $$ h(x)=\frac{f(x)-f(a)}{x-a} \\ g(x)=x-a $$ We are trying to find the limit: $$ \lim _{x \rightarrow a}[f(x)-f(a)] = \lim _{x \rightarrow a}[h(x)g(x)] $$ Using the product rule for limits, which states that if limit of both \(h(x)\) and \(g(x)\) exists then: $$ \lim_{x \rightarrow a}[h(x)g(x)] = \lim_{x \rightarrow a}h(x) \times \lim_{x \rightarrow a}g(x) $$

Step 4: Find limits of h(x) and g(x)

Now, we'll find the limits of \(h(x)\) and \(g(x)\) as x approaches a: 1. Since \(f'(a)\) exists, we already know the limit of \(h(x)\): $$ \lim_{x \rightarrow a}h(x) = f'(a) $$ 2. For \(g(x)\): $$ \lim_{x \rightarrow a}g(x) = \lim_{x \rightarrow a}(x-a) = a-a = 0 $$

Step 5: Apply the product rule for limits

Now, substituting these limits into the product rule for limits: $$ \lim _{x \rightarrow a}[f(x)-f(a)] = \lim _{x \rightarrow a}[h(x)g(x)] =f'(a) \times 0 = 0 $$

Step 6: Conclusion

Since the limit of \([f(x)-f(a)]\) as \(x\) approaches \(a\) is equal to 0, we have proved that if a function \(f\) is differentiable at \(x=a\), it must also be continuous at that point.

Key Concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the study of how things change. It provides tools for describing the rate at which quantities change, known as the derivative. In the context of a function, differentiability is a fundamental concept, as it signifies the existence of the derivative at a particular point. The derivative, denoted as \( f'(x) \), represents the slope of the tangent line to the function's graph at any given point, and thus gives us information about the function's instantaneous rate of change. In our exercise, the differentiability of the function \( f \) at \( x=a \) is demonstrated by showing that the limit of \( \frac{f(x)-f(a)}{x-a} \) exists as \( x \) approaches \( a \). This is a direct application of the concepts within differential calculus to understand and prove the behavior of functions and their continuity.

Limit of a Function

The limit of a function is a fundamental concept in calculus, capturing the idea of approaching a certain value. Mathematically, the expression \( \lim_{x \rightarrow a} f(x) \) represents the value that \( f(x) \) approaches as \( x \) gets arbitrarily close to \( a \). It's important to note that the function doesn't necessarily have to equal this value when \( x=a \); it only needs to approach it as \( x \) nears \( a \). In the problem we're discussing, the limit is used to show that as \( x \) approaches \( a \), the difference between \( f(x) \) and \( f(a) \) gets arbitrarily small, which implies that the function is continuous at that point. This concept is tightly linked with continuity and differentiability, as a function cannot be differentiable at a point if it isn't continuous there.

Product Rule for Limits

When dealing with the limits of products of functions, we employ the product rule for limits. According to this rule, if the limits of two functions \( h(x) \) and \( g(x) \) exist independently, then the limit of their product is the product of their limits: \( \lim_{x \rightarrow a}[h(x)g(x)] = \lim_{x \rightarrow a}h(x) \times \lim_{x \rightarrow a}g(x) \). In the step-by-step solution, this rule is applied to show that as \( x \) approaches \( a \), the product \( h(x)g(x) \), which corresponds to \( f(x) - f(a) \), approaches zero. This is because one of the factors, \( g(x) \), approaches zero, making the whole product zero and demonstrating the continuity of \( f \) at \( x=a \). The product rule for limits is a powerful tool for simplifying complex limit calculations.

Definition of a Derivative

The derivative of a function at a point provides the slope of the tangent to the function's curve at that point. It can be formally defined as the limit of the difference quotient as \( x \) tends to \( a \): \( f'(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} \). This expression gives the instantaneous rate of change of \( f(x) \) at \( x = a \). In the exercise, differentiability at a point is used to examine the behavior of the function near that point, and to establish that the function is continuous. If this limit (the definition of the derivative) does not exist, then the function is not differentiable at that point. The definition of a derivative is cornerstoned in differential calculus, and understanding it is essential for analyzing and interpreting the behavior of functions.