Prove $f:\mathbb{R^n}\to \mathbb{R}$ if differentiable is continuous

Question

Below I have written a proof for the problem in my title and was wondering if the proof works

$$0=\lim_{h\to 0}|f(a+h)-f(a)-\triangledown f\cdot h|$$

$$=\lim_{h\to 0}|f(a+h)-f(a)-\big(f_xh_1+\cdots f_nh_n\big)|$$

and since $h \to 0$ implies $(h_1,\cdots ,h_n) \Longrightarrow (0,\cdots , 0)$

$$\big(f_xh_1+\cdots f_nh_n\big) =0 \text{ as } h\to 0$$

So we get continuity since

$$0=\lim_{h\to 0}|f(a+h)-f(a)|$$

Answer 1

Here's a proof from the book (also one in English :)).

\begin{align*} \lim_{x \to x_0} f(x) = f(x_0) &\iff \lim_{x \to x_0} |f(x)- f(x_0)| = 0 \\ \\ &\iff \lim_{x \to x_0} \left|f(x)- f(x_0)\right| \frac{\|x-x_0\|}{\|x-x_0\|}= 0 \\ \\ & \iff \lim_{x \to x_0} \frac{|f(x)-f(x_0)|}{\|x-x_0\|}\|x-x_0\| = 0 \\ \\ & \iff \lim_{x \to x_0} \left(\frac{|f(x)-f(x_0)-f'(x_0)(x-x_0) + f'(x_0)(x-x_0)|}{\|x-x_0\|}\right)\|x-x_0\| = 0 \\ \\ & \leq \lim_{x \to x_0} \left(\frac{|f(x)-f(x_0)-f'(x_0)(x-x_0)| + |f'(x_0)(x-x_0)|}{\|x-x_0\|}\right)\|x-x_0\| \\ \\ & = \lim_{x \to x_0} \frac{|f(x)-f(x_0)-f'(x_0)(x-x_0)|}{\|x-x_0\|} \|x-x_0\| + |f'(x_0)(x-x_0)| \end{align*}

By the differentiability assumption we have,

$$\lim_{x \to x_0} \frac{|f(x)-f(x_0)-f'(x_0)(x-x_0)|}{\|x-x_0\|} = 0$$

Hence, each limit exists separately i.e;

\begin{align*} &\lim_{x \to x_0} \frac{|f(x)-f(x_0)-f'(x_0)(x-x_0)|}{\|x-x_0\|} \|x-x_0\| + |f'(x_0)(x-x_0)| \\ \\ & =\lim_{x \to x_0} \frac{|f(x)-f(x_0)-f'(x_0)(x-x_0)|}{\|x-x_0\|} \|x-x_0\| + \lim_{x \to x_0} |f'(x_0)(x-x_0)| \\ \\ & = 0 \end{align*}

Therefore,

$$ \lim_{x \to x_0} |f(x)-f(x_0)| = 0 \Rightarrow \lim_{x \to x_0} f(x) = f(x_0)$$

Answer 2

Probar que si $f:\mathbb{R^{n}}\longrightarrow \mathbb{R^{n}}$ es diferenciable en $a\in\mathbb{R^{n}}$, entonces es continua en $a$.

Supongamos que $f$ es diferenciable en $a\in\mathbb{R^{n}}$, es decir, existe una única transformación lineal $\lambda:\mathbb{R^{n}}\longrightarrow\mathbb{R^{n}}$ tal que $$ \displaystyle\lim_{h\longrightarrow 0} \frac{\vert f(a+h)-f(a)-\lambda (h)\vert}{\vert h\vert}=0. $$ Observe que $$ \begin{array}{cccccc} 0 & \leq & \vert f(a+h)-f(a)\vert & = & \left \vert \lambda(h)+\vert h\vert \cdot \dfrac{f(a+h)-f(a)-\lambda (h)}{\vert h\vert}\right\vert \\ & & & \leq & \vert\lambda (h)\vert +\vert h\vert\cdot\dfrac{\vert f(a+h)-f(a)-\lambda (h)\vert}{\vert h\vert} \\ & & & \leq & M\cdot\vert h\vert +\vert h\vert\cdot\dfrac{\vert f(a+h)-f(a)-\lambda (h)\vert}{\vert h\vert} \end{array} $$ Así, $$ \displaystyle\lim_{h\longrightarrow 0} 0\leq\displaystyle\lim_{h\longrightarrow 0} \vert f(a+h)-f(a)\vert\leq\displaystyle\lim_{h\longrightarrow 0} \vert h\vert\left(\vert\lambda\vert+\dfrac{\vert f(a+h)-f(a)-\lambda (h)\vert}{\vert h\vert}\right) $$

$ \begin{array}{rl} \Rightarrow & \displaystyle\lim_{h\longrightarrow 0} \vert f(a+h)-f(a)\vert=0 \\ & x=a+h \qquad h\longrightarrow 0 \Leftrightarrow x\longrightarrow a \\ \Rightarrow & \displaystyle\lim_{x\longrightarrow a} \vert f(x)-f(a)\vert=0 \\ \Rightarrow & \displaystyle\lim_{x\longrightarrow a} f(x)=f(a) \\ \Rightarrow & f \mbox { es continua en } a \end{array} $