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If Differentiable, Then Continuous 📂Analysis

If Differentiable, Then Continuous

Theorem1

Let’s say f:[a,b]Rf : [a,b] \to \mathbb{R}. If ff is differentiable at p[a,b]p \in [a,b], then ff is continuous at pp.

Explanation

Note that the converse ‘if it is continuous, it is differentiable’ does not hold.

In the past, there was a pun called Simple Integration (simply put, if it’s differentiable, then it’s continuous) among older students, but I wonder if this pun has become unused as current students might not know who Simple Integration is.

Proof

The equivalence condition for ff to be continuous at pp is as follows:

limxpf(x)=f(p) \lim \limits_{x \to p}f(x)=f(p)

Therefore, it suffices to show that limxp(f(x)f(p))=0\lim \limits_{x \to p} \left( f(x) - f(p) \right) = 0. Assume that ff is differentiable at pp. Then the following holds:

limxp(f(x)f(p))=limxp[f(x)f(p)xp(xp)]=limxpf(x)f(p)xplimxp(xp)=f(p)0=0 \begin{align*} \lim \limits_{x \to p} \left( f(x)-f(p) \right) &= \lim \limits_{x \to p}\left[ \frac{ f(x)-f(p) } {x-p}(x-p)\right] \\ &= \lim \limits_{x \to p} \frac{ f(x)-f(p) }{x-p} \cdot \lim \limits_{x \to p} (x-p) \\ &= f^{\prime}(p)\cdot 0 \\ &=0 \end{align*}

The second equality holds due to the properties of the limit of a function.


  1. Walter Rudin, Principles of Mathematical Analysis (3rd Edition, 1976), p104 ↩︎