# Math Help - Limit of A Sequence

1. Originally Posted by RockHard
Can anyone provide me with another example before I deem this thread solved
Prove that $\lim_{n\to\infty}\left(\sqrt{n^2+1}-n\right)=0$

2. Here's a trickier one.

Prove $\lim_{n\to\infty}n^2e^{-n}=0$

Hint: Rewrite $n^2e^{-n}$ as $e^{\ln(n^2)}e^{-n}=e^{2\ln(n)-n}$ and then try to put an upper bound on $2\ln(n)$ (that's easy to work with).

Note that if you follow the hint, your value for $N$ will look something like $N=\max\{c, f(\epsilon)\}$ for some number $c$ and some expression depending on $\epsilon$.

I hope what I wrote wasn't too confusing.

3. Whew, thanks man! Ill write these down and get to it, those look like a solid a example to work with, I hope I can get them correct or close to it.

4. The second one is (I think) significantly more difficult than the first, so don't worry if you have a hard time.

5. Not gonna lie I am a tad confused with the first one because of the extra term in the equation I know I am probalby over complicating it

6. Originally Posted by RockHard
Not gonna lie I am a tad confused with the first one because of the extra term in the equation I know I am probalby over complicating it
Try multiplying by $\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}$ to end up with $\frac{1}{\sqrt{n^2+1}+n}$.

7. I know this is a dumb question but I have to be sure before I attempt the problem, is everything under the square root or is it just n^2 + 1 also isnt it - n or I missed something in your hint

8. Originally Posted by RockHard
I know this is a dumb question but I have to be sure before I attempt the problem, is everything under the square root or is it just n^2 + 1 also isnt it - n or I missed something in your hint
Just $n^2+1$.

$\sqrt{n^2+1}-n\cdot\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n}=\frac{ n^2+1-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}$

9. Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

$\frac{1}{\sqrt{n^2+1}+n} < \epsilon$

$1 < {\epsilon}{{\sqrt{n^2+1}+n}}$

Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

Could I even do

${\sqrt{n^2+1}+n} > \frac{1}{\epsilon}$

Edit: I must continue so my erroneous ways can be corrected, even if it hurts, not literally.

${n^2+1+n^2} > ({\frac{1}{\epsilon}})^2$

${2n^2+1} > ({\frac{1}{\epsilon}})^2$

${2n^2} > ({\frac{1}{\epsilon}})^2 - 1$

Stuck

10. Originally Posted by RockHard
Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

$\frac{1}{\sqrt{n^2+1}+n} < \epsilon$

$1 < {\epsilon}{{\sqrt{n^2+1}+n}}$

Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

Could I even do

${\sqrt{n^2+1}+n} > \frac{1}{\epsilon}$
Merely notice that $\frac{1}{\sqrt{n^2+1}+n}\le\frac{1}{n}$
So letting $n>\frac{1}{\varepsilon}$ does the trick.

11. Originally Posted by RockHard
Man, my algebra never has been the greatest and I always get stuck on some problems that should be relatively easy because I second guess what I do if it is the correct way or not also, I know this horrible but what was the algebraic reason for multiplying by what you did? to make it fit the condition of absolute value a bit easier? but ill show a bit of work to where I am at

$\frac{1}{\sqrt{n^2+1}+n} < \epsilon$

$1 < {\epsilon}{{\sqrt{n^2+1}+n}}$

Usually around steps like these I cannot determine If I am doing it correctly and make silly mistakes

Could I even do

${\sqrt{n^2+1}+n} > \frac{1}{\epsilon}$

Edit: I must continue so my erroneous ways can be corrected, even if it hurts, not literally.

${n^2+1+n^2} > ({\frac{1}{\epsilon}})^2$

${2n^2+1} > ({\frac{1}{\epsilon}})^2$

${2n^2} > ({\frac{1}{\epsilon}})^2 - 1$

Stuck
What Drexel did was what I had in mind for the solution. If you do it this, way, you're going to get something like:

$\sqrt{n^2+1}+n>\frac{1}{\epsilon}\implies 2n^2+1{\color{red}+2n\sqrt{n^2+1}}>(1/\epsilon)^2\implies$ $2n\sqrt{n^2+1}>(1/\epsilon)^2-1-2n^2 \implies 4n^2(n^2+1)>[(1/\epsilon)^2-1-2n^2]^2$

and solving that equation for $n$ will give you a nasty quadratic.

12. Hmm, trying to make sense of what Drexel did, it is like it magically disappeared that nasty square root.

13. Originally Posted by RockHard
Hmm, trying to make sense of what Drexel did, it is like it magically disappeared that nasty square root.
All you have to realize is that saying $\frac{1}{\text{whatever}}<\frac{1}{\text{whatever else}}$ assuming positivity is that $\text{whatever}>\text{whatever else}$. Now just note that $n^2+1>n^2\implies \sqrt{n^2+1}>\sqrt{n^2}=n\implies \sqrt{n^2+1}+n>2n\implies \sqrt{n^2+1}+n>n$

14. So you can you always do this sort of thing, if it easy to find? What theorem is this called? There so many theorems and idea's you dont know when to apply what

15. Originally Posted by RockHard
So you can you always do this sort of thing, if it easy to find? What theorem is this called? There so many theorems and idea's you dont know when to apply what
Theorem? This is just a property of the real numbers since they are an ordered field? There isn't a theorem here, it's just the way it is.

Page 2 of 3 First 123 Last