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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by TheAbstractionist
Hi chella182.

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

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

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