1. ## Sequences - Limits

Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence ${a_n}$ if it exists. Otherwise explain why the sequence diverges.

(a) $a_n$ = $\frac{log n + 5n^2}{2n^2 + 100}$

I divided the whole thing with $n^2$ and using standard limits I got,

$\frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

Then I used l'Hopital's rule to solve the limit for the $\frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.

(b) $a_n = \frac{3^n + n!}{100^n + n^7}$

I don't know how to do this one!

(c) $a_n = (2^n + 1)^\frac{1}{n}$

This one is to use sandwich rule right?

(d) $a_n = cos(\frac {\pi n}{3n + 5})$

Used continuity rule and got,

$cos(\frac {lim \pi n}{lim 3n + 5})$
$cos(\frac { \pi lim n}{3 lim n + lim 5})$

divided the whole thing by n

$cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
$cos(\frac {\pi}{3})$
= 0.5

(e) $a_n = n tan(\frac{1}{n})$

I don't know how to do this one either.

2. Hello

Originally Posted by pearlyc
Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence ${a_n}$ if it exists. Otherwise explain why the sequence diverges.

(a) $a_n$ = $\frac{log n + 5n^2}{2n^2 + 100}$

I divided the whole thing with $n^2$ and using standard limits I got,

$\frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

Then I used l'Hopital's rule to solve the limit for the $\frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.
L'Hôpital's rule would yield :

$\frac{\frac 1n}{2n}=\frac{1}{2n^2}$, and the limit of this is 0.

don't havd time for thinking about the following ones, sorry

3. Originally Posted by pearlyc
(d) $a_n = cos(\frac {\pi n}{3n + 5})$

Used continuity rule and got,

$cos(\frac {lim \pi n}{lim 3n + 5})$
$cos(\frac { \pi lim n}{3 lim n + lim 5})$

divided the whole thing by n

$cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
$cos(\frac {\pi}{3})$
= 0.5
This is right

I will answer the remaining in a while..

4. Originally Posted by pearlyc
Okay, I attempted 3 of these and I don't know how to do the other two. So I hope I can get you guys to check if I am on the right track and teach me how to do the other two. Thanks a lot, a lot, a lot!

Use the airthmetic of limits, standard limits (clearly stated) or appropriate rules (clearly stated) to compute the limit of each sequence ${a_n}$ if it exists. Otherwise explain why the sequence diverges.

(a) $a_n$ = $\frac{log n + 5n^2}{2n^2 + 100}$

I divided the whole thing with $n^2$ and using standard limits I got,

$\frac{\lim\infty{\frac{log n}{n^2}} + 5}{2}$

Then I used l'Hopital's rule to solve the limit for the $\frac{log n}{n^2}$ and got 1/2 and just substituted it back in and my answer is 11/4.

(b) $a_n = \frac{3^n + n!}{100^n + n^7}$

I don't know how to do this one!

(c) $a_n = (2^n + 1)^\frac{1}{n}$

This one is to use sandwich rule right?

(d) $a_n = cos(\frac {\pi n}{3n + 5})$

Used continuity rule and got,

$cos(\frac {lim \pi n}{lim 3n + 5})$
$cos(\frac { \pi lim n}{3 lim n + lim 5})$

divided the whole thing by n

$cos(\frac { \pi lim 1}{3 lim 1 + lim \frac{5}{n}})$
$cos(\frac {\pi}{3})$
= 0.5

(e) $a_n = n tan(\frac{1}{n})$

I don't know how to do this one either.
For b)

Remember that the factorial function grows faster than polynomials or exponentials. The series diverges .

For c)

try this trick

$a_n = (2^n + 1)^\frac{1}{n}$ Take the natural log of both sides

$\ln(a_n) = \ln \left((2^n + 1)^\frac{1}{n}\right))$

using log properties we get

$\ln(a_n)=\frac{1}{n} \cdot \ln(2^n+1) =\frac{\ln(2^n+1)}{n}$
We can now use L'hospitals rule to get

$\ln(a_n)=\frac{\frac{(\ln(2))2^n}{2^n+1}}{1}$

Now letting n go to infinity gives

$\ln(a_n)=\ln(2) \iff a_n=2$

For d you are correct

for e) rewrite as $\frac{\tan(\frac{1}{n})}{\frac{1}{n}}$

and use L.H rule

Good luck.

5. Hi
(c) $a_n = (2^n + 1)^\frac{1}{n}$

This one is to use sandwich rule right?
Yes, you can use the squeeze theorem : $2^{n+1} > 2^n+1 > 2^n$ hence ...

6. Whoa, thanks for the many responses guys (: Took your guidance and attempted the questions!

For (c), this is how far I've got ..

$\sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}$
$lim (2)^\frac{1}{n} < \sqrt[n]{2^n + 1} < lim 2.2^\frac{1}{n}$
$1 < \sqrt [n]{2^n+1} < 2$

Where do I go from here? Hmm.

As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

After differentiating, $\frac {-1}{x^2} sec^2x$

I don't know where to go from here too

7. Originally Posted by pearlyc
$\sqrt[n]{2^n} <\sqrt[n]{2^n + 1} < \sqrt[n]{2^{n+1}}$
$lim (2)^\frac{1}{n} < \sqrt[n]{2^n + 1} < lim 2.2^\frac{1}{n}$

You missed a minor point.
Try again... what is $\sqrt[n]{2^n}$

As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

After differentiating, $\frac {-1}{x^2} sec^2x$

I don't know where to go from here too
LH rule can be a little dangerous here. Instead modify the question a little bit and see if you can recognize the limit...

When $n \to \infty, \frac1{n} \to 0$, so lets call $\frac1{n}$ as $\theta$.

Now where have I seen $\lim_{\theta \to 0} \frac{\tan \theta}{\theta}$?

If you havent seen this limit before, LH is still an option on this...

8. Oh OOPS!

Thanks, hahaha.

Eh, I still don't really get that tan question!

9. Originally Posted by pearlyc
Oh OOPS!

Thanks, hahaha.

Eh, I still don't really get that tan question!
Limit(n to infinity) is equivalent to

Try L'Hospitals rule

10. Originally Posted by pearlyc
As for (e), I followed TheEmptySet's advice and used L.H. rule, and this is what I've got,

After differentiating, $\frac {-1}{x^2} sec^2x$

I don't know where to go from here too
Well part of the problem is you shouldn't have ended up here.

Note that the orginial sequence was

$n\tan\left( \frac{1}{n}\right)$

Rewriting as $\frac{\tan\left( \frac{1}{n}\right)}{\frac{1}{n}}$

as $n \to \infty$ this goes to $\frac{0}{0}$

Now applying L'hospitials rule we get

$\frac{\sec^{2}(\frac{1}{n})\cdot (\frac{-1}{n^2})}{(\frac{-1}{n^2})}$

This is where your error occured you forgot to use the chain rule when taking the derivative of the tangent function

Now when we reduce we get

$\lim_{n \to \infty}{\sec^{2}\left( \frac{1}{n}\right)} \to \sec^2(0)=1^2=1$

I hope this clears it up.

Good luck.