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One-sided limits for the difference quotient of |x| at 0

Consider the absolute value function f(x)=∣x∣f(x) = |x|.

Using the limit definition f′(0)=lim⁡h→0f(0+h)−f(0)hf'(0) = \lim_{h \to 0} \dfrac{f(0+h) - f(0)}{h}, compute the one-sided limits lim⁡h→0+∣h∣h\lim_{h \to 0^+} \dfrac{|h|}{h} and lim⁡h→0−∣h∣h\lim_{h \to 0^-} \dfrac{|h|}{h} separately, and use these to state whether f′(0)f'(0) exists. Give the value of each one-sided limit, and explain your conclusion about differentiability of f(x)=∣x∣f(x) = |x| at x=0x=0.

Hint

Recall that ∣h∣=h|h| = h when h>0h>0 and ∣h∣=−h|h| = -h when h<0h<0. Evaluate ∣h∣h\frac{|h|}{h} separately in each case, then compare the two results to decide if the two-sided limit exists.

Step-by-step solution

For h>0h > 0, ∣h∣=h|h| = h, so f(0+h)−f(0)h=hh=1\frac{f(0+h)-f(0)}{h} = \frac{h}{h} = 1 for all such hh. Therefore lim⁡h→0+∣h∣h=1\lim_{h \to 0^+} \frac{|h|}{h} = 1.

For h<0h < 0, ∣h∣=−h|h| = -h, so f(0+h)−f(0)h=−hh=−1\frac{f(0+h)-f(0)}{h} = \frac{-h}{h} = -1 for all such hh. Therefore lim⁡h→0−∣h∣h=−1\lim_{h \to 0^-} \frac{|h|}{h} = -1. Since the one-sided limits (11 and −1-1) disagree, the two-sided limit lim⁡h→0∣h∣h\lim_{h\to 0}\frac{|h|}{h} does not exist, so f′(0)f'(0) does not exist and f(x)=∣x∣f(x)=|x| is not differentiable at x=0x=0 (this corresponds to the corner in its graph).

Sign of the two one-sided limit values

Hint
Step-by-step solution

The right-hand limit equals one, which is positive, while the left-hand limit equals negative one, which is negative. So the right-hand value is positive and the left-hand value is negative. Because these two values differ, the overall two-sided limit does not exist, confirming that f prime of zero does not exist for the absolute value function.

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