The limit functional on the sequence space of all convergent sequences is bounded functional with norm 1 Let $c$ be the set of all convergent real (or complex) sequences and $l_\infty $ be the set of all bounded real (or complex) sequences. Now $l_\infty$ is a normed linear space with respect to the well known sup-norm defined by $$\|(x_n)_n\|_\infty=\sup_n |x_n|$$
Then, we can define a linear functional $f:c\to \mathbb K$ by $$f((x_n)_n)=\lim_{n\to \infty} x_n$$ where, $\mathbb K=\mathbb R$ or $\mathbb C$
I've read that $\|f\|=1.$
My approach : $$|f(x)|=|\lim_n x_n|\leq \sup_n |x_n|=\|x\|_\infty$$ Thus, $$\frac{|f(x)|}{\|x\|_\infty}\leq 1$$ Which implies that $\|f\|\leq1$.
My Question : How can I show that $\|f\|\geq1$ ?

Hint: Consider a constant sequence.
