Differentiability implies continuity ($f(x+h)-f(x)=h[f^{\prime }(x)+o(1)]$)

If a real function has a finite derivative at $x$, its difference quotient tends to $f^{\prime }(x)$ and is bounded for sufficiently small nonzero increments. Multiplying it by the increment shows $f(x+h)-f(x)\to 0$, proving continuity at $x$. The converse fails: the absolute value function is continuous but not differentiable at zero.