11.21 - More about slant asymptotes Consider the hyperbola...

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More about slant asymptotes: Consider the hyperbola x 2 y 2 = 1: - 2 - 1 0 1 2 - 2 - 1 0 1 2 To see that the hyperbola x 2 y 2 = 1 has the line y = x as a slant asymptote, take the upper branch y = sqrt( x 2 – 1) and show that it gets arbitrarily close to the line y = x as x + : “rationalizing the numerator”, we get lim x + x – sqrt( x 2 – 1) = lim x + 1/( x + sqrt( x 2 – 1)) = 0. You can likewise show that the line y = – x is also a slant asymptote. Indeed, both lines are two-sided asymptotes. Problem 4.4.49: xy = x 2 + 4 y = x + 4/ x y x = 4/ x , which goes to 0 as x + and as x . So y = x is a two-sided slant asymptote.
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- 10 - 5 5 10 - 10 - 5 5 10 Problem 4.4.50: y = e x x y – (– x ) = y + x = e x , which goes to 0 as x . So y = – x is a one-sided slant asymptote. - 3 - 2 - 1 1 2 3 - 3 - 2 - 1 1 2 3 [Digression: How do we show that e x x goes to infinity as x + ? I gave a proof in class, but here’s a better way that doesn’t require integration: Write e x x = e x (1 – x / e x ). We know
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11.21 - More about slant asymptotes Consider the hyperbola...

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