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Unformatted text preview: nce.
a. Point at the equation
xn
and then click on the button nn n
n! e n to supply the definition to Scientific Notebook. When you see the screen make the selection “A function argument” so that Scientific Notebook knows that you are defining a
sequence.
b. Point at the expression x n and click on the button to display the sequence graphically. Revise your graph and set the domain interval as 1, 500 . Double click into your graph to make the
buttons appear in the top right corner and click on the bottom button to select it. Trace your graph with the
mouse and show graphically that 161 lim nÝ c. nn n
0. 3989.
n! e n Point at the expression
nn n
n! e n
and ask Scientific Notebook to evaluate it numerically. Compare the result with the limit value you found
graphically.
lim
nÝ d. Point at the expression and ask Scientific Notebook to evaluate it exactly to show that the limit is
1/ 2 . 2. Prove that 5 n /n! 0 as n Ý. Solution: Whenever n 5 we have
5n 5
5
5
5
0
1
2
3
4
n!
Now, to prove that 5 n /n! 0 as n Ý, suppose that
Choose an integer N such that 54
4! 5
55
n
5
6
0. 5
n 5
N 5 .
4! Then whenever n N we have
5n
n! 0
3. Prove that n!/n n 0 as n 54
4! 54
4! 5
n 5
N . Ý. Hint: Make use of the fact that, for each n we have
0 n!
nn 1
n 2n
n
n 1.
n 4. Given that x n is a sequence of real numbers, that x 0 and that x n x as n
integer N such that the inequality x n 0 holds for all integers n N. Hint: The interval Ý, prove that there exists an 0, Ý is a neighborhood of the number x. 5. Given that x n 0 for every positive integer n and that x is a partial limit of the sequence x n , prove that
x 0.
We need to show that no negative number can be a partial limit of x n . Suppose that y 0. Since
the interval Ý, 0 is a neighborhood of y and x n is not frequently (or, indeed, ever) in the interval
Ý, 0 , we deduce that y is not a partial limit of x n .
6. Suppose that x n is a sequence of real numbers and that x
equivalent:
a. x n x as n R. Prove that the following conditions are Ý. b. x n x  0 as n Ý.
Condition a says that for every 0 there exists an integer N such that whenever n
have x n x  .
Condition b says that for every 0 there exists an integer N such that whenever n
have x n x  0  .
These two conditions clearly say the same thing.
7. Suppose that x n is a sequence of real numbers, that x
n Ý. 162 R and that x n x as n N we
N we Ý. Prove that x n  x  as Hint: Make use of the fact that for each n we have
0 x n  x  x n x . 8. Suppose that x n is a sequence of real numbers, that x R and that x n x as n Ý. Suppose that p is an
integer and that for every positive integer n we have
y n x n p .
Prove that y n x as n Ý.
Suppose that 0. Using the fact that x n x as n Ý we choose an integer N such that
whenever n N we have x n x  . We observe that the inequality y n x  will hold whenever
n p N which is the same as saying that n N p.
9. Given that a n b n for every positive integer...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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