Equality between the two limits We have a function: $\phi : U \to \mathbb{R}^l$, where $U \subseteq \mathbb{R}^k$ is an open subset. $\phi$ is a function of class $C^1$, differentiable in $q$, $q \in U$. Let $\displaystyle q_n \to q$ and $\displaystyle \phi(q_n) \to \phi(q)$ while $\displaystyle n \to \infty$. How to prove that:
$$ \lim_{n \to \infty} \frac{D\phi (q) (q_n - q)}{||q_n -q||} = \lim_{n \to \infty} \frac{\phi (q_n) - \phi(q)}{||q_n -q||} \text{?} $$ There is a hint to use the definition of $D\phi(q)$ and a fact that it is a linear map. I tried to write it using the definition, but it didn't work. Please, help me. 
 A: It is not true in general.
Let $\phi(x) = x$, and choose $q=0$, and $q_n = (-1)^n\frac{1}{n}$. Then $\frac{q_n-q}{\|q_n-q\|} = (-1)^n$, and hence $\frac{D\phi(q)(q_n-q)}{\|q_n-q\|} = (-1)^n$, which has no limit as $n \to \infty$.
What is true, however, is that $\lim_n (\frac{\phi(q_n)-\phi(q)}{\|q_n-q\|} - \frac{D\phi(q)(q_n-q)}{\|q_n-q\|}) = 0$.
To see this:
By definition of $D\phi(q)$, you have for all $\epsilon>0$, there exists a $\delta>0$ such that if $\|q'-q\|< \delta$, then $\|\phi(q')-\phi(q) - D\phi(q)(q'-q)\| \leq \epsilon \|q'-q\|$.
Let $\epsilon >0$, then for $n$ sufficiently large we have $\|q_n-q\| < \delta$, and so $\|\phi(q_n)-\phi(q) - D\phi(q)(q_n-q)\| \leq \epsilon \|q_n-q\|$.
If we assume that $q_n \neq q$ for all $n$ sufficiently large, we can divide across to get $\|\frac{\phi(q_n)-\phi(q)}{\|q_n-q\|} - \frac{D\phi(q)(q_n-q)}{\|q_n-q\|}\| \leq \epsilon$, from which it follows that $\lim_n (\frac{\phi(q_n)-\phi(q)}{\|q_n-q\|} - \frac{D\phi(q)(q_n-q)}{\|q_n-q\|}) = 0$.
