Limits using standard results

[atsoc15]

Having a few problems getting my head around soling limits.

Basically solutions and methods to this will help me on my way.

Un = (1-10n)/n+10

find the limit as n tends to infinity.

thanks in advance

[atsoc15]

think i might have actually esolved this. would i be right and saying that
1-10n/n+10


= (1/n-10)/1+10/n)
which as n tends to infinity the limit is -10/1 being -10.

Is this correct

[opus]

Whats the correct formula?

-A- (1-10n)/(n+10)

or
-B- (1-10n)/n +10 which is the same as ((1-10n)/n)+10 ??

Neiter one is corresponding to the formula in your second post

= (1/n-10)/1+10/n)

(at least one bracket is missing!)

the limit for -A- is -10,
for -B- it would be 0!

[atsoc15]

the exact formula is (1-10n) / (n+10)

i think ive got the hang of doing this using standard methods. but im struggling to do this vigorously. Im assuming a use the sandwich theorem but its a much more complex problem. any chance u could show us what to do for it.

[opus]

I don't have to slightest idea what you're talking about!
I solved your question like this:

(1-10n)/(n+10)

first look at (1-10n), or -10n +1 , with large n the difference to 10n can be neglected. I.E. take it as 10n

than look at (n+10), also for large n it's can be viewed as n

so the formula would be cloes to -10n/n, which is -10!

And where is your sandwich? (Peanutbutter/jelly ???)

[atsoc15]

i may be mistaken but the theory you used being similar to mine i think is called waving hands or something like that. in other words its very rough and not enough for a mathematical prrof. another method being the sandwich theorem where u find the closest limit above and below then work it out from there. prob better described here
http://en.wikipedia.org/wiki/Sandwich_theorem

[zaza]

How advanced do you want to be?

Split the formula on the numerator:

f(n) = [1/(n+10)] - [10n/(n+10)]

and let n -> inf.

The first part clearly -> 0

As for the second part, it is more difficult to tell because the top and bottom will both end up at inf.
Use L'Hopital's Rule, which states that lim f(x)/g(x) = lim f'(x)/g'(x) and differentiate top and bottom, then do the limit....

...we get -10.

So the limit as n-> inf is -10.


zaza

[atsoc15]

ok what about this then (10) + ((10sin10n)/n) as n tends to infinity

[zaza]

You need to be careful when sending trig functions to infinity, because they are not single valued. But in principle you can solve it just as I did above for your previous question, particularly if you're looking at n-> 0 instead.

If you didn't understand something I posted before, feel free to ask questions. But despite all appearances, this isn't a homework forum.


zaza