0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.