MATH220 April 2005
• Q1 • Q2 • Q3 (a) • Q3 (b) • Q3 (c) • Q4 • Q5 • Q6 • Q7 • Q8 • Q9 • Q10 • Q11 •
Prove, directly from the -definition of the limit of a sequence, that
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A sequence converges to a number L if for any there exists a number such that
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Let ε > 0, we want to show that for some positive integer Nε, we have that
First, we can get rid of the absolute value since
and so we wonder if for some value of n large enough, we will have
for the value of ε that we fixed (and that we think of as being a very small number). Since we are interested in the value of n, we take the logarithm on each side and obtain
Now since 2/3 is a number smaller than 1, its logarithm will be a negative number, so when we divide in the above by ln(2/3) we have to flip the inequality sign, we obtain
Note that since ε is also a very small number, its logarithm is also negative and hence the fraction on the right-hand side is a positive number (as expected). We can thus choose Nε to be the nearest integer larger than the fraction, which we can write as
which concludes our proof.
It is not a bad idea to see that ε proofs allow you to actually compute something about how quickly the sequence will converge to its limit. In this case for example, if we choose ε = 0.001 then the above proof gives us the value
So for any n ≥ 18, the value of (-2/3)n should be less than 0.001, indeed we have
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