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Help with Limits problem

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  1. Dec 14, 2016 #1
    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. jcsd
  3. Dec 14, 2016 #2
    Take a look at your second term ##\frac 7 {3n^2+7}##, what does this approach as n goes to infinity?
     
  4. Dec 14, 2016 #3
    Zero?
     
  5. Dec 14, 2016 #4
    Yes. Since it's effectively ##\frac 1 \infty##. Now, what does that leave you with?
     
  6. Dec 14, 2016 #5
    (1)^(n^2+1) ?
     
  7. Dec 14, 2016 #6
    So, what is 1 to any power?
     
  8. Dec 14, 2016 #7
    1 I think, sorry Im a bit slow today.
     
  9. Dec 14, 2016 #8
    Yep. One multiplied by itself infinite times will get you one.

    Edit: Ignore, turned out to be incorrect.
     
    Last edited: Dec 14, 2016
  10. Dec 14, 2016 #9
  11. Dec 14, 2016 #10
    Assuming you didn't write the limit "as n goes to 8" in that last line on purpose, yes.
     
  12. Dec 14, 2016 #11
    Woops yeah that would be infinity, thanks a lot.
     
  13. Dec 14, 2016 #12
    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
  14. Dec 14, 2016 #13

    Mark44

    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.

    @TJGilb, please take care when you give advice that your advice is factually correct.
     
  15. Dec 14, 2016 #14
    Thanks glad I noticed the correct version. Appreciate it, this community is amazing!
     
  16. Dec 14, 2016 #15
    Don't be so happy, you won't get full solutions like this from next time. :(((
     
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