If $g:R^m \to R^n$ has derivative $\lambda$ at $a$, is the limit as $x\to 0$ of $\frac{|g(a + x) - g(a)|}{|x|}$ always $\|\lambda\|$? I am able to show the limit is bounded above by $\|\lambda\|$ if it exists: $$\frac{|g(a +x) - g(a)|}{|x|} \leq \frac{|g(a+ x) - g(a) - \lambda x|}{|x|} + \|\lambda\|$$
 A: If you mean $g:\mathbb{R} \rightarrow \mathbb{R}^n$, the answer is yes. Since:
$$\lim_{x \rightarrow 0}\left|\left|\frac{g(x + a) - g(a)}{x} - \lambda \right|\right| = 0$$
Now we can use the squeeze theorem. First a lemma: For any vectors $a$ and $b$ in $\mathbb{R}^n$, $-\|a - b\| \leq \|a\| - \|b\| \leq \|a - b\|$. You can prove this using the triangle inequality: $\|a - b + b\| \leq \|a - b\| + \|b\|$, so $\|a\| - \|b\| \leq \|a - b\|$, this is the right inequality. The left inequality is given by switching $a$ and $b$: $\|b\| - \|a\| \leq \|b - a\|$, then multiply by $-1$. 
So:
$$-\left|\left|\frac{g(x + a) - g(a)}{x} - \lambda \right|\right| \leq \frac{\|g(x + a) - g(a)\|}{\|x\|} - \|\lambda\| \leq \left|\left|\frac{g(x + a) - g(a)}{x} - \lambda \right|\right|$$
The squeeze theorem implies that the middle goes to zero as well.
Now what if $g:\mathbb{R}^n \rightarrow \mathbb{R}^m$? A simple case is $g:\mathbb{R}^2 \rightarrow \mathbb{R}$, $g(u,v) = v$. Then setting $a = 0$:
$$\frac{\|g(u,v) - g(0,0)\|}{\|(u,v)\|} = \frac{|v|}{\sqrt{u^2 + v^2}}$$
And this limit does not exist as $(u,v) \rightarrow (0,0)$. 
A: That limit hardly ever exists. For example, take $g:\mathbb {R}^2\to \mathbb {R}$ defined by $g(x,y) = x.$ Then
$$|g((0,0)+(x,y)) - g((0,0))|/(x^2+y^2)^{1/2} = |x|/(x^2+y^2)^{1/2},$$
which doesn't have a limit as $(x,y)\to (0,0).$
My guess is the limit exists iff $\lambda $ is a scalar multiple of a linear isometry.
