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One-sided limits of the difference quotient at a corner point

Consider the function f(x)=∣x−3∣f(x) = |x - 3|. We want to investigate whether ff has a tangent line at x=3x = 3 by examining the difference quotient that defines the derivative.

Using the limit definition f′(3)=lim⁡h→0f(3+h)−f(3)h,f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}, compute the one-sided limits as h→0+h \to 0^+ and h→0−h \to 0^- separately. Based on these two one-sided limits, does f′(3)f'(3) exist, and is f(x)=∣x−3∣f(x) = |x-3| differentiable at x=3x = 3? Explain your conclusion about whether a tangent line exists at x=3x=3.

One-sided limits of the difference quotient at a corner point graph
Hint

Write f(3+h)=∣3+h−3∣=∣h∣f(3+h) = |3+h-3| = |h|, and recall that ∣h∣=h|h| = h when h>0h > 0 but ∣h∣=−h|h| = -h when h<0h < 0. Compute the difference quotient separately for each sign of hh and compare the two resulting limits.

Step-by-step solution

Since f(3)=∣3−3∣=0f(3) = |3-3| = 0 and f(3+h)=∣h∣f(3+h) = |h|, the difference quotient is ∣h∣h\frac{|h|}{h}. For h→0+h \to 0^+, we have ∣h∣=h|h| = h, so ∣h∣h=hh=1\frac{|h|}{h} = \frac{h}{h} = 1, giving a right-hand limit of 11.

For h→0−h \to 0^-, we have ∣h∣=−h|h| = -h, so ∣h∣h=−hh=−1\frac{|h|}{h} = \frac{-h}{h} = -1, giving a left-hand limit of −1-1. Since the one-sided limits (11 and −1-1) are not equal, lim⁡h→0∣h∣h\lim_{h\to 0} \frac{|h|}{h} does not exist, so f′(3)f'(3) does not exist. Therefore f(x)=∣x−3∣f(x) = |x-3| is not differentiable at x=3x=3, and there is no tangent line there — the graph has a corner at that point.

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