For particularly, $\epsilon-N$ proofs I was wondering why the following proposition holds true.
Let $a,b\in \mathbb{R}$. Then, $\displaystyle{\lim_{n\rightarrow \infty}}a+b\cdot f(n)=a+b\cdot \displaystyle{\lim_{n\rightarrow \infty}}f(n)$.
For particularly, $\epsilon-N$ proofs I was wondering why the following proposition holds true.
Let $a,b\in \mathbb{R}$. Then, $\displaystyle{\lim_{n\rightarrow \infty}}a+b\cdot f(n)=a+b\cdot \displaystyle{\lim_{n\rightarrow \infty}}f(n)$.
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Let $l = \lim\limits_{n\to +\infty} f(n)$.
For any $\epsilon > 0, \exists N \in \mathbb{N}\text{ } / \text{ }n \geq N \Rightarrow |f(n)-l| \leq \epsilon$.
Then, for $ n \geq N$,
$$ |(a+b f(n))-(a+bl)| = |b|.|f(n)-l| \leq |b|\epsilon.$$