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Unformatted text preview: nd it using our example f (x) = 2x+11 . If we interpret f (x) as a division problem, (2x−1)÷(x+1),
we ﬁnd the quotient is 2 with a remainder of −3. Using what we know about polynomial division,
speciﬁcally Theorem 3.4, we get 2x − 1 = 2(x + 1) − 3. Dividing both sides by (x + 1) gives
x+1 = 2 − x+1 . (You may remember this as the formula for g (x) in Example 4.1.1.) As x becomes
unbounded in either direction, the quantity x+1 gets closer and closer to zero so that the values of
f (x) become closer and closer to 2. In symbols, as x → ±∞, f (x) → 2, and we have the result.7
Notice that the graph gets close to the same y value as x → −∞ or x → ∞. This means that
the graph can have only one horizontal asymptote if it is going to have one at all. Thus we were
justiﬁed in using ‘the’ in the previous theorem. (By the way, using long division to determine the
asymptote will serve us well in the next section so you might want to review that topic.)
Alternatively, we can use what we know...
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