Question

Who is really good at explaining how to do limits?

Answer 1

Don't worry about me, I'm just here to watch... out of interest, do you have a specific example?

Answer 2

whoever deleted that thanks. and yeah hold on @terenzreignz

Answer 3

were example?

Answer 4

when it is shown as 1+ (-1)^n/n (and the limit being n-->infinity) how would you do something like that

Answer 5

Ambiguity kills... is it this limit
\[\huge \lim_{n\rightarrow \infty}1+\frac{(-1)^n}{n}\]

Answer 6

yeah pretty much...she said the study guide is basically the test...im looking at my notes and it just doesnt make sense :/

Answer 7

Well, thankfully, the limit of a sum is the sum of the limits IF both limits exist...
In maths language
\[\Large \lim_{n\rightarrow \infty}(a_n+b_n)=\lim_{n\rightarrow \infty}a_n +\lim_{n\rightarrow \infty} b_n\]

Again, only if both limits exist.

Answer 8

exactly so cant i write it (sorry idk how to really use the equation thing for these) lim 1+lim(-1)^n/n...but what will that do?

Answer 9

Well, clearly, this limit
\[\large \lim_{n\rightarrow \infty}1\]exists, right?

So it would be lovely (LOL) if this limit
\[\Large \color{blue}{\lim_{n\rightarrow \infty}\frac{(-1)^n}{n}}\]also existed... now does it?

Answer 10

Actually, it does, do you suspect as much? :)
Where do you think this sequence converges?

Answer 11

i think it doe, im not even sure. limits and me are just bad...

Answer 12

Okay, notice that as n gets bigger and bigger, the (-1)^n just keeps alternating from positive and negative, but the denominator gets bigger and bigger too.

That means, regardless of its sign, the absolute value of this part \(\Large \frac{(-1)^n}{n}\) gets smaller and smaller.

Right? :)

Answer 13

converge is when it approaches a limit. i know that.

Answer 14

right!

Answer 15

So... you wanna go for the full definition of a limit?
Or do you accept that the limit of \(\Large \frac{(-1)^n}{n}\) as n goes to infinity is zero? :)

Answer 16

i accept!

Answer 17

i have to find the limit..so wouldnt it be zero? it cant be that easy can it!? \-_-/

Answer 18

So.. since clearly
\[\huge \lim_{n \rightarrow \infty} 1 = 1 \]and
\[\huge \lim_{n \rightarrow \infty} \frac{(-1)^n}{n}=0\]

then what's the limit
\[\huge \lim_{n\rightarrow \infty}1+\Large \frac{(-1)^n}{n}\]?

Answer 19

1!

Answer 20

Remember... the limit of a sum is the sum of the limits if both limits exist, and both limits do, in fact, exist.

And you're correct :)
The answer is indeed, 1.
Nicely done :D

Answer 21

omg i love you so much right now! THANK YOU THANK YOU THANK YOUI!

Answer 22

No problem :)