## Calculus: Early Transcendentals 8th Edition

This is not true. The counterexample is the function $f(x)=|x-a|.$ It is continuous everywhere but at the point $x=a$ there is a problem because there the function has two different tangents with two different slopes: $$m_1=\lim_{h\to 0^-}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0^-}\frac{|a-a+h|-|a-a|}{h}=\lim_{h\to0^-}\frac{|h|-0}{h}=\lim_{h\to0^-}\frac{-h}{h}=\lim_{h\to0^-}-1=-1.$$ $$m_1=\lim_{h\to 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0^+}\frac{|a-a+h|-|a-a|}{h}=\lim_{h\to0^+}\frac{|h|-0}{h}=\lim_{h\to0^+}\frac{h}{h}=\lim_{h\to0^+}1=1.$$ This means that the limit defining the derivative $$\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ does not exist so the function is not differentiable at $x=a$.