# Thread: Proving that lim(k + a_{n}) = k + lim(a_{n})

1. ## Proving that lim(k + a_{n}) = k + lim(a_{n})

Sorry for the naff title, don't really know how else to put it. The question goes...

Let $a_{n}$ be a sequence tending to $a$ and let $k$ be a real number. Give an $\epsilon-N$ proof that $\lim_{n\rightarrow\infty}(k+a_{n})=k+\lim_{n\right arrow\infty}a_{n}$.

Looks simple enough, but I can't get a handle on it at all.

2. Originally Posted by chella182
Sorry for the naff title, don't really know how else to put it. The question goes...

Let $a_{n}$ be a sequence tending to $a$ and let $k$ be a real number. Give an $\epsilon-N$ proof that $\lim_{n\rightarrow\infty}(k+a_{n})=k+\lim_{n\right arrow\infty}a_{n}$.

Looks simple enough, but I can't get a handle on it at all.
Hi chella182.

Given $\epsilon>0,\ a_n\to a\ \implies$ there is $N$ such that $|a_n-a|<\epsilon$ for all $n>N.$ The result follows immediately as $\left|(k+a_n)-(k+a)\right| = \left|a_n-a\right|.$

3. In general if we have two sequence $a_{n}$ and $b_{n}$ with...

$\lim_{n \rightarrow \infty} a_{n}=a$ , $\lim_{n \rightarrow \infty} b_{n}=b$

... and $f(x,y)$ is a continuos function $\forall [x,y] \in \mathbb{R}^{2}$ is...

$\lim_{n \rightarrow \infty} f(a_{n},b_{n}) = \lim_{n \rightarrow \infty} f(a_{n}, b) = \lim_{n \rightarrow \infty} f(a,b_{n}) = f(a,b)$

In your case is $f(x, y) = x + y$ ...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by TheAbstractionist
Hi chella182.

Given $\epsilon>0,\ a_n\to a\ \implies$ there is $N$ such that $|a_n-a|<\epsilon$ for all $n>N.$ The result follows immediately as $\left|(k+a_n)-(k+a)\right| = \left|a_n-a\right|.$
Is that it? That the solution? :S

5. Originally Posted by chella182
Is that it? That the solution? :S
Hi chella182.

Yes. Let $b_n=k+a_n$ and $b=k+a.$ Then you want that $\forall\,\epsilon>0,$ you can find some $N$ such that $n>N\ \implies\ \left|b_n-b\right|<\epsilon,$ right? That was what I proved.