# Limits to e!Watch

#1
Okay, I understand that the lim(n->inf) (1+(1/n))^n = e
but I was just wondering what these limits where..

lim(n->inf) (2+(1/n))^n = ??

lim(n->inf) (1+(5/n))^n = ?? e^5n ??

I guess it's just trivial, but any help is great!

Thanks
0
10 years ago
#2
First one would, for obvious reasons, be infinite, as (2+1/n)^n > 2^n. Second would be e^5, which can be proved by a similar method to the fact that e = lim (1 + 1/n)^n.
0
10 years ago
#3
(Original post by sleeping in)
Okay, I understand that the lim(n->inf) (1+(1/n))^n = e
but I was just wondering what these limits where..

lim(n->inf) (2+(1/n))^n = ??

lim(n->inf) (1+(5/n))^n = ?? e^5n ??

I guess it's just trivial, but any help is great!

Thanks
really? or do you just believe it
0
#4
So what about ((3/5)+(1/n))^n as n->inf

thanks.
0
10 years ago
#5
(Original post by sleeping in)
So what about ((3/5)+(1/n))^n as n->inf

thanks.
What do you think?

(What happens when n > 5?)
0
#6
I'm guessing it will tend to zero then?
0
10 years ago
#7
(Original post by sleeping in)
I'm guessing it will tend to zero then?
Correct. Why?
0
#8
Well for n<5, we have terms greater than 1, however when we start looking at n>5 our 1/n gets progressively smaller and as we don't have '1+' are expression will get smaller and smaller, as soon (3/5 + 1/n) << 1.

Now, if we are looking at limits of sequences, can we use the same argument? considering for example,

(a)_n>1 = (3/5 + (1/n))^n

_ denotes subscript,

Thanks
0
10 years ago
#9
(Original post by sleeping in)
Well for n<5, we have terms greater than 1, however when we start looking at n>5 our 1/n gets progressively smaller and as we don't have '1+' are expression will get smaller and smaller, as soon (3/5 + 1/n) << 1.
Well, basically. For n > 5, we have 1/n < 1/5 and so 0 < (3/5 + 1/n)^n < (4/5)^n which converges to 0 (geometric progression). Thus (3/5 + 1/n)^n converges (comparison test) to 0 (squeeze / sandwich theorem / other silly names).

(Original post by sleeping in)
Now, if we are looking at limits of sequences, can we use the same argument? considering for example,

(a)_n>1 = (3/5 + (1/n))^n

_ denotes subscript,

Thanks
You mean you want to find where ? Yes, same argument - the limit's 0.
0
#10
(Original post by generalebriety)
Well, basically. For n > 5, we have 1/n < 1/5 and so 0 < (3/5 + 1/n)^n < (4/5)^n which converges to 0 (geometric progression). Thus (3/5 + 1/n)^n converges (comparison test) to 0 (squeeze / sandwich theorem / other silly names).

You mean you want to find where ? Yes, same argument - the limit's 0.
That's great, thanks a lot!
0
10 years ago
#11
No problem.
0
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