Problem 43 - Calculus Volume 1

Question

Use the limit definition of derivative to show that the derivative does not exist at \(x=a\) for each of the given functions. $$ f(x)=\left\\{\begin{array}{l} 1, x<1 \\ x, x \geq 1 \end{array}, x=1\right. $$

Step-by-Step Solution

Answer

The derivative does not exist at \( x = 1 \) because the left and right limits are different (0 and 1, respectively).

Step 1: Identify the Limit Definition of Derivative

The derivative of a function \( f(x) \) at a point \( x = a \) is defined as \( f'(a) = \lim_{h\to0}\frac{f(a+h)-f(a)}{h} \), provided this limit exists.

Step 2: Determine \( f(a) \)

Since we want to find if the derivative exists at \( x = 1 \), we need to know \( f(1) \). From the function definition, when \( x = 1 \), we have \( f(1) = 1 \).

Step 3: Evaluate Right-hand Limit

To find the right-hand limit, consider \( h > 0 \). We have \( f(1+h) = 1+h \) as \( 1+h \geq 1 \).Calculate the difference quotient: \[ \lim_{h\to0^+} \frac{(1+h) - 1}{h} = \lim_{h\to0^+} \frac{h}{h} = \lim_{h\to0^+} 1 = 1. \]

Step 4: Evaluate Left-hand Limit

For \( h < 0 \), we have \( f(1+h) = 1 \) since \( 1+h < 1 \) in this case.The difference quotient is: \[ \lim_{h\to0^-} \frac{1 - 1}{h} = \lim_{h\to0^-} 0 = 0. \]

Step 5: Compare Right and Left Limits

The right-hand limit as \( h \to 0^+ \) is 1, while the left-hand limit as \( h \to 0^- \) is 0. Since these two limits are not equal, the overall limit does not exist at \( x = 1 \).

Key Concepts

Piecewise Functions

A piecewise function is defined in different ways over different intervals. Each section of the function has its own rule or expression. In the given exercise, the function \( f(x) \) is defined piecewise:\

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• For \( x < 1 \), \( f(x) = 1 \)
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• For \( x \geq 1 \), \( f(x) = x \)
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\This means the function behaves differently depending on whether \( x \) is less than 1 or greater than or equal to 1. Understanding piecewise functions involves recognizing these shifts and how they change the function's behavior. It is important to learn how to work with each piece in its own domain. This can affect calculations related to limits and derivatives, especially at key points where the definition changes, like at \( x = 1 \) in this exercise.

Right-hand Limit and Left-hand Limit

When dealing with limits of piecewise functions, we often compare a function's approach from two directions: the right and the left. The right-hand limit, denoted as \( \lim_{h \to 0^+} \), involves approaching a point from values greater than the point.
For the given function at \( x = 1 \), we calculate the right-hand limit for \( f(x) \) as \( h \rightarrow 0^+ \). Since \( f(1 + h) = 1 + h \), it simplifies to \( \lim_{h \to 0^+} 1 = 1 \).

Meanwhile, the left-hand limit, denoted as \( \lim_{h \to 0^-} \), focuses on approaching the point from lesser values. For \( x = 1 \), \( f(1 - h) = 1 \), leading to \( \lim_{h \to 0^-} 0 = 0 \).

These limits are essential for determining the continuity and differentiability of a function at a certain point. In this case, because the right-hand and left-hand limits are not equal at \( x = 1 \), the overall limit does not exist there. This difference is crucial for understanding the function's behavior at transition points.

Existence of Derivatives

The existence of a derivative at a point depends on the function having a single limit as it approaches the point from both directions.
Using the limit definition of the derivative, \( f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \), both left-hand and right-hand limits should be the same.

In the given function, we observe two different behaviors: \( \lim_{h \to 0^+} 1 = 1 \) and \( \lim_{h \to 0^-} 0 = 0 \).
This discrepancy indicates that \( f'(1) \) does not exist, as the function is not differentiable at \( x = 1 \).

For a derivative to exist, continuity and smoothness around the point are vital. Since this piecewise function lacks such conditions at the transition point, the derivative fails to exist. Such understanding is critical in calculus to determine where and when a function is differentiable.