Example 1
Show that
lim
x® 1
3x+5 = 8.
lim
x® 2
x2+3x = 10.
lim
x® 3
1
x
=
1
3
.
Example 5
Show that
lim
x® ¥
Öx = ¥.
Let E be given. We can assume that E ³ 0 , because if it weren't, you could replace it by the larger number 0. Now, let's find the number D . But how, you might ask, can we find D if we don't know what E is, really? The idea is to find a D expressed in terms of E that works, so that, no matter what E is, D is adjusted automatically. You figure that out by a sort of reverse-engineering: In order to make Öx > E , try to see if you can get an inequality involving x from the one involving f(x) . Here the way to do it is to square both sided of the inequality. Since everything in the inequality is positive, that won't change the inequality (you have to worry about that sort of thing, sometimes). Then: Öx > EÛ x > E2,
so, if x > E2 , we'll guarantee that Öx > E . So, E2 will do as our D , D = E2 .
Note that I said that that choice of D will do. There isn't any one answer here, you just need to find a way to get a large enough x ; larger will also do nicely.
Example 6
Show that
lim
x® ¥
(x/2+sinx) = ¥.
lim
x® ¥
2x+3
x-1
= 2.
Assume that e > 0 is given. We need to be able to get
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Then,
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which means that
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or, solving for x , and taking x > 0 (since x® ¥,
this is not a severe restriction):
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By David L. Johnson, last modified 2/22/00.