陳述：如果函式 f ( x ) {\displaystyle f(x)} 在點 x 0 , {\displaystyle x_{0},} 可導，則 f {\displaystyle f} 在 x 0 . {\displaystyle x_{0}.} 連續。
證明：假設 f ( x ) {\displaystyle f(x)} 在 x 0 {\displaystyle x_{0}} 可導。則 D x f ( x 0 ) {\displaystyle D_{x}f(x_{0})} 存在。
函式 f ( x ) {\displaystyle f(x)} 在 x 0 {\displaystyle x_{0}} 連續，當 lim x → x 0 f ( x ) = f ( x 0 ) . {\displaystyle \lim _{x\to x_{0}}f(x)=f(x_{0}).} ⇒ lim x → x 0 f ( x ) − f ( x 0 ) = 0 {\displaystyle \Rightarrow \lim _{x\to x_{0}}f(x)-f(x_{0})=0} ⇒ lim x → x 0 f ( x ) − lim x → x 0 f ( x 0 ) = 0 {\displaystyle \Rightarrow \lim _{x\to x_{0}}f(x)-\lim _{x\to x_{0}}f(x_{0})=0} ⇒ lim x → x 0 [ f ( x ) − f ( x 0 ) ] = 0. {\displaystyle \Rightarrow \lim _{x\to x_{0}}[f(x)-f(x_{0})]=0.} 因此，為了證明該定理，只需證明 lim x → x 0 [ f ( x ) − f ( x 0 ) ] = 0. {\displaystyle \lim _{x\to x_{0}}[f(x)-f(x_{0})]=0.} 。
lim x → x 0 [ f ( x ) − f ( x 0 ) ] {\displaystyle \lim _{x\to x_{0}}[f(x)-f(x_{0})]} = lim x → x 0 [ f ( x ) − f ( x 0 ) x − x 0 ( x − x 0 ) ] {\displaystyle =\lim _{x\to x_{0}}[{\frac {f(x)-f(x_{0})}{x-x_{0}}}(x-x_{0})]} = lim x → x 0 [ f ( x ) − f ( x 0 ) x − x 0 ] ⋅ lim x → x 0 ( x − x 0 ) {\displaystyle =\lim _{x\to x_{0}}[{\frac {f(x)-f(x_{0})}{x-x_{0}}}]\cdot \lim _{x\to x_{0}}(x-x_{0})} = D x f ( x 0 ) ⋅ 0 = 0 {\displaystyle =D_{x}f(x_{0})\cdot 0=0}
因此，當 D x f ( x 0 ) {\displaystyle D_{x}f(x_{0})} 存在時， lim x → x 0 [ f ( x ) − f ( x 0 ) ] = 0 {\displaystyle \lim _{x\to x_{0}}[f(x)-f(x_{0})]=0} 成立。
∴ {\displaystyle \therefore } 如果 f ( x ) {\displaystyle f(x)} 在點 x 0 , {\displaystyle x_{0},} 可導，則 f {\displaystyle f} 在 x 0 . {\displaystyle x_{0}.} 連續。[證明完畢]
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