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My question is the following: If we have $q_n \to q$, $f(q_n) \to f(q)$, $f : \mathbb{R}^k \to \mathbb{R}^l$ is differentiable (so continuous), do we have $$\lim_{n \to \infty} \frac{\displaystyle \lim_{n \to \infty} f(q_n) - f(q)}{||q_n - q||} = \lim_{n \to \infty} \frac{ f(q_n) - f(q)}{||q_n - q||}?$$ Or do we need some extra assumptions?

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If $\lim_{n\to\infty}f(q_n)=f(q)$, then the numerator of the limit in the LHS will be $0$ and thus the LHS too will be $0$. –  Sayantan Jan 27 '13 at 13:16
    
I think what the OP is asking is basically why the RHS limit exists. –  Git Gud Jan 27 '13 at 13:19
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Try $f:\mathbb R^2\to\mathbb R$ defined by $f(x,y)=\color{red}{5}x+\color{blue}{3}y$, with $q_{2n}=(1/n,0)$ and $q_{2n-1}=(0,1/n)$, hence $q_n\to q=(0,0)$, and use the shorthand $$ g_n=\frac{f(q_{n})-f(q)}{\|q_{n}-q\|}. $$ Then $f$ is $C^\infty$ everywhere but $f(q)=0$, $f(q_{2n})=\color{red}{5}/n$, $f(q_{2n-1})=\color{blue}{3}/n$ and $\|q_n-q\|\sim2/n$ hence $g_{2n}\to\color{red}{5}$ while $g_{2n-1}\to\color{blue}{3}$. Thus the sequence $(g_n)_{n\geqslant1}$ diverges.

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