Let ff continous and and (xn)n∈N(x_{n})_{n\in \mathbb{N}} a sequence bounded. If limn→âˆ‍=alim_{n\rightarrow âˆ‍}=a, prove that \left \| f(a) \right \|\leq c\left \| f(a) \right \|\leq c

Let f:\mathbb{R}^{m}\rightarrow \mathbb{R}^{p}f:\mathbb{R}^{m}\rightarrow \mathbb{R}^{p} be a continous function and (x_{n})_{n\in \mathbb{N}}(x_{n})_{n\in \mathbb{N}} a sequence in \mathbb{R}^{m}\mathbb{R}^{m} with \left \| f(x_{n}) \right \|\leq c\left \| f(x_{n}) \right \|\leq c for every n\in \mathbb{N}n\in \mathbb{N}. If lim_{n\rightarrow âˆ‍}=alim_{n\rightarrow âˆ‍}=a, prove that \left \| f(a) \right \|\leq c\left \| f(a) \right \|\leq c

My attempt

(x_{n})_{n\in \mathbb{N}}(x_{n})_{n\in \mathbb{N}} is convergent, then:

For every \epsilon>0, \exists N\in \mathbb{N}\epsilon>0, \exists N\in \mathbb{N} such that \left \| x_{n} -a \right \| < \epsilon\left \| x_{n} -a \right \| < \epsilon for every n \geq \mathbb{N}n \geq \mathbb{N}. ff is continous, then: For every \epsilon>0, \exists \delta>0\epsilon>0, \exists \delta>0 such that \left \| f(x_{n}) -f(a) \right \|< \epsilon\left \| f(x_{n}) -f(a) \right \|< \epsilon, when \left \|x_{n} -a \right \|<\delta\left \|x_{n} -a \right \|<\delta Taking \epsilon=c+\left \| f(x_{n}) \right \| >0\epsilon=c+\left \| f(x_{n}) \right \| >0 and by triangle inequality

\left | \left \| f(x_{n}) \right \|-\left \| f(a) \right \| \right | \leq \left \| f(x_{n}) -f(a) \right \|< c+\left \| f(x_{n}) \right \|\left | \left \| f(x_{n}) \right \|-\left \| f(a) \right \| \right | \leq \left \| f(x_{n}) -f(a) \right \|< c+\left \| f(x_{n}) \right \| \left | \left \| f(x_{n}) \right \|-\left \| f(a) \right \| \right | \left \| f(a) \right \| > -c-2\left \| f(x_{n}) \right \|c>\left \| f(a) \right \| > -c-2\left \| f(x_{n}) \right \|

So we have that \left \| f(a) \right \|