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Example 3
In the limit
find a real number B > 0 such that whenever t > B, (2 + 3/t) is strictly within 1/100 of 2. To find B, we assume t > B and t > 0, and get inequalities for 2 + 3/t.
Now choose B so that 3/B < 1/100. The number B = 300 will do. It follows that whenever t > 300,
and
is strictly within
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Last Update: 2006-11-15