The Student Room Group

limit

https://www.dropbox.com/s/dq9lm6rkv8rf5ry/ex.gif

I have used the Lagrange s theorem,but I don t know how to continue after.
Original post by laur21
https://www.dropbox.com/s/dq9lm6rkv8rf5ry/ex.gif

I have used the Lagrange s theorem,but I don t know how to continue after.

It's unbounded. That's kind of obvious by asymptotic arguments - n goes to infinity, en/nen+1/(n+1)e^n/n - e^{n+1}/(n+1) is roughly en/nen+1/ne^n/n-e^{n+1}/n as nn \to \infty, and that goes to -\infty.
My method was a bit more systematic - I expressed ennen+1n+1\frac{e^n}{n}-\frac{e^{n+1}}{n+1} as an integral, and showed that the integral tends to -infinity.
Reply 2
the expression inside is equal to:

en(1e1+1n) e^n(1-\frac{e}{1+\frac{1}{n}})
Reply 3
Original post by jassi1
the expression inside is equal to:

en(1e1+1n) e^n(1-\frac{e}{1+\frac{1}{n}})

How is that going to help me solving this exercise?
(edited 9 years ago)
Original post by laur21
How is that going to help me solving this exercise?

Look carefully at the sign of 1e1-e.
Reply 5
Original post by Smaug123
Look carefully at the sign of 1e1-e.

Oh right..I was expecting a more elegant and spectacular way to solve this.
/dissapointed
Original post by laur21
Oh right..I was expecting a more elegant and spectacular way to solve this.
/dissapointed

My way was quite spectacular, I thought, if not elegant. (Integrate the bracket with respect to n.)
Reply 7
Original post by Smaug123
My way was quite spectacular, I thought, if not elegant. (Integrate the bracket with respect to n.)

Yes,it was :smile:
Thanks mate

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