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Chapter 5: Problem 10

Show that the absolute value function \(f(x):=|x|\) is continuous at every point \(c \in \mathbb{R}\).

Short Answer

Expert verified
The function \(f(x):=|x|\) is continuous at every point in the real line because for all \(c \in \mathbb{R}\), the limit as \(x\) approaches \(c\) of \( |x| \) equals \( |c| \).

Step by step solution

01

Set up The Definition of Continuity

Recall that the definition of a continuous function at a point is that: \[ \lim_{x \to c} f(x) = f(c) \] Apply this to the absolute value function to get: \[ \lim_{x \to c} |x| = |c| \] This means that as \(x\) approaches \(c\), the value of the function \(|x|\) should approach \(|c|\).
02

Evaluate the Limit

Since the real line splits into two regions by the absolute value function, consider the limit as \(x\) approaches \(c\) from the right and from the left separately: \[ \lim_{x \to c^+} |x| \quad and \quad \lim_{x \to c^-} |x| \] Since \(|x|\) is either \(x\) or \(-x\) depending on whether \(x\) is greater or less than 0, resolve the two cases accordingly. If \(c > 0\), then \( \lim_{x \to c^+} |x| = c \) and \( \lim_{x \to c^-} |x| = c \). Similarly, if \(c < 0\), we have \( \lim_{x \to c^+} |x| = -c \) and \( \lim_{x \to c^-} |x| = -c \). If \( c = 0 \), both limits equal 0.
03

Conclude Continuity

Confirm that for all \(c\), \( \lim_{x \to c} |x| = |c| \). Therefore, by the definition of continuity, \(|x|\) is continuous at every point \(c \in \mathbb{R}\).

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