First the graph of y f x certainly seems to possess

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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 2x−1 3 x+1 = 2 − x+1 . (You may remember this as the formula for g (x) in Example 4.1.1.) As x becomes 3 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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