find the limit of the sequence
n-1/n

- amyna

find the limit of the sequence
n-1/n

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- amyna

you take the limit as n approaches infinity

- amyna

but i don't know how to start

- zepdrix

Is it \(\large\rm \frac{n-1}{n}\) or \(\large\rm n-\frac{1}{n}\) ?

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## More answers

- amyna

the first one

- zepdrix

Oh limit of the sequence? not the series?
Ok that makes things a little easier then :d

- zepdrix

\[\large\rm \lim_{n\to\infty}\frac{n-1}{n}\quad=\quad \lim_{n\to\infty}\frac{n}{n}-\frac{1}{n}\quad=\quad \lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]You could break it up like this I suppose.
That seems like a good start :o

- amyna

ok i will try this approach, can i also multiply both sides by n ?

- zepdrix

No no let's not try anything fancy like that :)
We don't actually have any "sides" here.
It's just this guy \(\large\rm \lim\limits_{n\to\infty}\frac{n-1}{n}\)
and we're rewriting him in a way that's easier to handle.
Our limit depends on n,
So in the same way that you can't factor n `out of the limit`,
Example:\(\large\rm \lim\limits_{n\to\infty}2\frac{1}{n}\ne\frac{1}{n}\lim\limits_{n\to\infty}\frac{1}{n}2\) (Can't pull the n stuff out like that)
we also don't want to introduce new n stuff into the limit! :o

- zepdrix

Ah typo in my example :c woops

- zepdrix

I was trying to say that you can't factor the 1/n outside of the limit.
Was just trying to give an example why you don't want to multiply any n's into the limit.

- amyna

ok got it

- zepdrix

\[\large\rm =\quad \lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]But if we were able to break it down into these two limits,
these should be manageable.
You don't remember that second limit? :o

- amyna

then how would i solve it after i break the function up? sorry it has been long since i learned limits so i don't really remember how to actually solve it.

- zepdrix

I applied a limit rule to break it into two separate limits.
As `n approaches infinity`, what does the number 1 approach?

- amyna

infinity as well ?

- zepdrix

careful! we're not adding up a bunch of 1's.
We're just looking at that number 1 as n grows and grows.
1 is a constant.
It's unchanging.
So as n gets larger and larger,
1 approaches 1.
Hopefully that makes some sense :o

- zepdrix

But for the other part...
As `n approaches infinity`...
let's look at a really big n to see what's going on.\[\large\rm \lim_{n\to\infty}\frac{1}{n}\approx\frac{1}{12bajillion}\]I plugged in a really really big number for n,
letting it get "close" to infinity.

- zepdrix

This is one of those limits that you really want to remember :)
But oh well hehe.
So do you notice anything about this number?
How about a number you can plug into your calculator,
like 1/99999?

- amyna

so how would i show the necessary work for this problem since it approaches 1?

- zepdrix

You do these simple algebra steps,\[\large\rm \lim_{n\to\infty}\frac{n-1}{n}\quad=\quad \lim_{n\to\infty}\frac{n}{n}-\frac{1}{n}\quad=\quad \lim_{n\to\infty}1-\frac{1}{n}\]And then,
at this point in your class,
your teacher probably wants you to state what rule you're using here.
\[\large\rm =\lim_{n\to\infty}1-\lim_{n\to\infty}\frac{1}{n}\]Justify that step by saying something like...
`The limit of a difference is equivalent to the difference of limits`.
Or whatever the name of that limit rule is, I can't remember.
\[\large\rm =1-0\]\[\large\rm =1\]I think that's all you need to do :o
I don't think you need to say anything special about the 1/n limit.
That's a pretty common one.
I could be wrong though.

- amyna

Okay this makes so much more sense now! Thank You for your help!

- zepdrix

yay team \c:/

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