Solving Limits Problem: Struggling with Missed Lecture Material

[Bob Jonez]

Homework Statement

I was sick and missed the lecture so having hard time with this problem @_@.
http://i.imgur.com/ByK7iVk.png

Homework 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.

The Attempt at a Solution

My attempt and where i got stuck.
http://i.imgur.com/ONvliRo.jpg

 

[TJGilb]

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

 

[Bob Jonez]

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

Zero?

 

[TJGilb]

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

 

[Bob Jonez]

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

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

 

[TJGilb]

So, what is 1 to any power?

 

[Bob Jonez]

So, what is 1 to any power?

1 I think, sorry I am a bit slow today.

 

[TJGilb]

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

Edit: Ignore, turned out to be incorrect.

 

[Bob Jonez]

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

http://i.imgur.com/Wnmz4no.jpg

So that would be the whole soulution?

 

[TJGilb]

So that would be the whole soulution?

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

 

[Bob Jonez]

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

Woops yeah that would be infinity, thanks a lot.

 

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

 

[Mark44]

So, what is 1 to any power?

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

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.

@TJGilb, please take care when you give advice that your advice is factually correct.

 

[Bob Jonez]

@Bob Jonez, @TJGilb
I think the limit is not 1.

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

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

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

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

 

[Buffu]

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

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