1

Why does differentiability imply continuity?

For instance, consider $f(x)=x$ on $(-\infty,0]$ and $f(x)=x+1$ on $(0,\infty)$. Then the at 0 the left derivative equals the right derivative equals 1, so why doesn't $f'(0)=1$?

vukov
  • 1,555
  • Think of it this way: when a function (or interval of a function) is differentiable, you have an existing point on which you can draw a tangent. Hence, for that point to exist, it must be continuous. – Shahar Mar 07 '14 at 14:06
  • 2
    You are taking the limit of the derivatives, not that of the difference quotient at $0$. – alex Mar 07 '14 at 14:09

3 Answers3

1

The right derivative is: $$ f'^+(0)=\lim_{x,0^+}\frac{x+1-0}{x}=+\infty $$

7raiden7
  • 1,794
0

The right derivative at the origin is not $1$; it's undefined, since the line connecting the point $(h,f(h))=(h,h+1)$ (for $h>0$) to the point $(0,f(0))=(0,0)$ becomes infinitely steep as $h$ approaches zero.

Hans Lundmark
  • 53,395
0

Theorem. Let $f$ be differentiable at $x_0$. Then $f$ is continuous at $x_0$.

Yourr function cannot be differentiable, because $\lim_{x \to 0+} f(x) =1 \neq 0=f(0)$.

Siminore
  • 35,136