Let $f:I\to\mathbb{R}^n$ be a differentiable function, with $f'(a)\neq 0$ for some $a$ in the interval $I\subset\mathbb{R}$. If there exists a line $L\subset\mathbb{R}^n$ and a sequence $(x_k)$ in $I$such that $x_i\neq x_j$ when $i\neq j$, $\lim x_k=a$ and $f(x_k)\in L$ for all $k\in\mathbb{N}$, then $L$ is the tangent line to $f$ at point $a$.

This is what I've tried: the tangent line to $f$ at point $a$ is the set $T=\{f(a)+tf'(a);\;\;t\in\mathbb{R}\}$. So, it's needed to show that $L=T$. Suppose that $L=\{u+tv;\;\;t\in\mathbb{R}\}$ for some $u,v\in\mathbb{R}^n$. Then for all $k\in\mathbb{N}$ there exists $t_k\in \mathbb{R}$ such that $f(x_k)=u+t_kv$. Moreover there exists $t_a\in\mathbb{R}$ such that $f(a)=u+t_av$. Thus

$$f'(a)=\lim_{k\to \infty}\frac{f(x_k)-f(a)}{x_k-a}=\lim_{k\to \infty}\frac{(u+t_kv)-(u+t_av)}{x_k-a}=\lim_{k\to \infty}\left(0u+\frac{t_k-t_a}{x_k-a}v\right)$$

Since $f$ is differentiable, it's continuous. So, $\lim f(x_k)=f(a)$.

Therefore, we know that $f'(a),f(a)\in L$ (because $L$ is closed). Could someone give me a hint to finish?


You have most of the pieces, you just have to arrange them correctly. To use your language, we need to show that $L=\{u+tv;\;\;t\in\mathbb{R}\}$ satisfies $f(a)\in L$ and $v$ is parallel to $f'(a)$.

proof that $f(a)\in L$:

Note that $\lim_{k\to\infty}x_k=a$. Since $f$ is differentiable, it is continuous, which is to say that $\lim_{k\to\infty}f(x_k)=f(a)$. Since $f(x_k)\in L$ for each $k$, we know that $f(a)$ is a limit point of $L$. Since $L$ is closed, $f(a)\in L$. Thus, there is some $t_a$ so that $L(t_a)=f(a)$.

proof that $v$ is parallel to $f'(a)$:

As you stated, $$ f'(a)=\lim_{k\to \infty}\frac{f(x_k)-f(a)}{x_k-a} $$ However, since each $f(x_k)\in L$, we can also say that $$ \begin{align} L'(a)&=\lim_{k\to\infty}\frac{f(x_k)-f(a)}{t_k-t_a}\\ &=\lim_{k\to\infty}\frac{f(x_k)-f(a)}{x_k-a}\cdot\frac{x_k-a}{t_k-t_a}\\ &=f'(a)\cdot \lim_{k\to\infty} \frac{x_k-a}{t_k-t_a} \end{align} $$ Since $L'(t)=v$, we deduce that $v$ is a scalar multiple of $f'(a)$, which means that the two vectors are parallel.

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  • $\begingroup$ Where did you use $f'(a)\neq 0$? $\endgroup$ – Pedro Jul 25 '13 at 15:04
  • $\begingroup$ The vector equality implicitly does so. If $f'(a)$ were $0$, we wouldn't be able to say anything about the direction of $L'(a)$ $\endgroup$ – Ben Grossmann Jul 25 '13 at 15:45
  • $\begingroup$ Where did you use $x_i\neq x_j$ if $i\neq j$? $\endgroup$ – Pedro Jul 25 '13 at 16:06
  • $\begingroup$ I didn't. It is sufficient that $\lim_{k\to\infty}x_k=a$ $\endgroup$ – Ben Grossmann Jul 25 '13 at 16:30
  • $\begingroup$ How do you know that the sequence $(x_k-a)/(t_k-t_a)$ converges? $\endgroup$ – Pedro Jul 25 '13 at 16:43

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