Question

r=lim 1/2*lim(1+1/n)/1

NEXT STEP:
r=1/2*1+0/1
^^ How do you get that?

Answer 1

wat is this

Answer 2

pre calculus limits.

Answer 3

oh one minute I'll look it u

Answer 4

Just keep the following fact in mind:
limn!1
n
1=n = limn!1
e
1
n
logn = e
limn!1
1
n
logn = e
0 = 1
because
limx!1
1
x
log x = limx!1
log x
x
L’H= limx!1
1=x
1
= 0
We’ll also use Striling’s approximation:
p
2 n
n+1=2
e
n
n!
n
n+1=2
e
n1
for all n 2 N
Problem 1 (3.6).
(a) P1
n=1
n+1
n
Diverges: limn!1
n+1
n = 1 6= 0. In general, if P
n
an converges, then limn!1 an = 0 –
sometimes this is called the Divergence Test for some reason.
(f) P1
n=1
n
4
2n
Converges: lim sup

n
4
2n


1=n = limn!1
n
4=n
2 = limn!1
(n
1=n)
4
2 = 1=2 < 1. Convergence is
by the Root Test.
(g) P1
n=1
p
n
(n)n =
P1
n=1
(1)n
nn2
Converges: limn!1
1
nn2 = 0, so convergence is by the Alternating Series Test.
(h) P1
n=1
n!
nn
Converges: Observe that,
n!
nn
=
1 2 3 (n 1) n
nn
=
1
n

2
n

3
n

n 1
n

n
n
=
=
1
n

2
n


3
n

n 1
n

n
n

<
2
n2
Then, convergence is by the Comparison Test – comparing to P1
n=1
2
n2 , which converges by
the p-Test.
(i) P1
n=1 sin n
2
Diverges: limn!1 sin n
2
doesn’t exist – see (a).
(l) P1
n=1
n+2
n3+5
Converges: Could use the Root Test, but I’ll use the Comparison Test.
n + 2
n3 + 5
=
1
n3

1
n2 +
2
n3
1 + 5
n3

1
n3

3
6
=
1
2n3
and P1
n=1
1
2n3 converges by the p-Test.
1

Answer 5

wat happened i copy and paste and thats wat happened

Answer 6

@Hope_nicole write your equation clearly and legibly.

Answer 7

\[\lim \frac{ 1 }{ 2 }* \lim \frac{ 1+1/n }{ 1 }\]
the next step goes to this:\[\frac{ 1 }{ 2 }*\frac{ 1+0 }{ 1 }\]

Answer 8

i dont see how you get from the first step to the last step.

Answer 9

what the the limit to????
limits look like this:
\[\lim_{x\to a}f(x)\]

Answer 10

idk how to write it with the equation thing on here but it is n-->infinity.

Answer 11

ok. you see. that information is crucial.. otherwise, it is like saying, "I bought __les with __dol__"

Answer 12

so, is this like this?
\[\lim_{n\to\infty}{1\over2}\lim_{n\to\infty}1+{1\over n}\]

Answer 13

no the last one is\[\frac{ 1+1/n }{ 1}\]

Answer 14

ok..
remember that anything divided by "1" gives you the same thing

Answer 15

\[1\times x=x\implies x={x\over 1}\]

Answer 16

you limit
\[
L=\left(\lim_{n\to\infty}{1\over2}\right)\left(\lim_{n\to\infty}1+{1\over n}\right)\\
\]
the first paranthesis is independent of "n" so, it will not change.
and in the second one, we subtitute the limit
\[L={1\over2}\left(1+{1\over\infty}\right)\\
\text{but}\quad{1\over\infty}=0\\
L={1\over2}(1+0)={1\over2}\]