# Help with Limits problem

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1. Dec 14, 2016

### Bob Jonez

1. The problem statement, all variables and given/known data
I was sick and missed the lecture so having hard time with this problem @_@.
http://i.imgur.com/ByK7iVk.png

2. Relevant equations
I don't know how to solve it all the textbook don't have particular problem so having a hard time figuring it out.

3. The attempt at a solution
My attempt and where i got stuck.
http://i.imgur.com/ONvliRo.jpg

2. Dec 14, 2016

### TJGilb

Take a look at your second term $\frac 7 {3n^2+7}$, what does this approach as n goes to infinity?

3. Dec 14, 2016

### Bob Jonez

Zero?

4. Dec 14, 2016

### TJGilb

Yes. Since it's effectively $\frac 1 \infty$. Now, what does that leave you with?

5. Dec 14, 2016

### Bob Jonez

(1)^(n^2+1) ?

6. Dec 14, 2016

### TJGilb

So, what is 1 to any power?

7. Dec 14, 2016

### Bob Jonez

1 I think, sorry Im a bit slow today.

8. Dec 14, 2016

### TJGilb

Yep. One multiplied by itself infinite times will get you one.

Edit: Ignore, turned out to be incorrect.

Last edited: Dec 14, 2016
9. Dec 14, 2016

### Bob Jonez

10. Dec 14, 2016

### TJGilb

Assuming you didn't write the limit "as n goes to 8" in that last line on purpose, yes.

11. Dec 14, 2016

### Bob Jonez

Woops yeah that would be infinity, thanks a lot.

12. Dec 14, 2016

### Buffu

Mentor note: This member has been warned about posting complete solutions.
$$\lim_{n \to \infty} \left({3n^2\over 3n^2 + 7} \right)^{n^2 + 1} = \lim_{n \to \infty} \left(1 +{-7\over 3n^2 + 7} \right)^{3n^2 + 3 + 4 - 4\over 3}$$

$$\lim_{n \to \infty} \left(1 +{-7\over 3n^2 + 7} \right)^{3n^2 + 7 - 4\over 3} =\lim_{n \to \infty} {\left(1 +{-7\over 3n^2 + 7} \right)^{3n^2 + 7\over 3} \over \left(1 +{-7\over 3n^2 + 7} \right)^{4 \over 3}} = \lim_{n \to \infty} {\left(1 +{-7/3\over (3n^2 + 7)/3} \right)^{3n^2 + 7\over 3}}$$

Now using $e^x = \lim_{n \to \infty} \left(1+ {x\over n}\right)^n$
the given limit is
$$\lim_{n \to \infty} {\left(1 +{-7/3\over (3n^2 + 7)/3} \right)^{3n^2 + 7\over 3}} = e^{-7 \over 3}$$

Last edited by a moderator: Dec 14, 2016
13. Dec 14, 2016

### Staff: Mentor

This is not true, in general. For example, $\lim_{n \to \infty}(1 + 1/n)^n = e$. Even though the base is approaching 1, and the exponent is becoming unbounded, the limit is not equal to 1, a contradiction to what you said above.

14. Dec 14, 2016

### Bob Jonez

Thanks glad I noticed the correct version. Appreciate it, this community is amazing!

15. Dec 14, 2016

### Buffu

Don't be so happy, you won't get full solutions like this from next time. :(((