2.6 - Review Key

X 2 x 2 1x 2x2 00 lim lim 0 2 x 10 2 x

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Unformatted text preview: (x − 2)3 x→2 x→2 0 3 3 3 Since lim = lim , we conclude that lim does not exist 3 3 x→2 (x − 2)3 x→2+ (x − 2) x→2− (x − 2) (DNE). 9. lim 10. Let f (x) = x2 − x − 6 . (3x + 5)(x2 − 5x + 6) (a) Identify all vertical and horizontal asymptotes of f (x) by giving their equations. To find the vertical asymptotes we simplify and look for values where the denominator of f(x) is 0. So, (x − 3)(x + 2) (x + 2) x2 − x − 6 = = (3x + 5)(x2 − 5x + 6) (3x + 5)(x − 2)(x − 3) (3x + 5)(x − 2) as long as x = 3. Setting the denominator equal to 0 and solving for x we have: (3x + 5)(x − 2) = 0 ⇒ 3x + 5 = 0 or x − 2 = 0 5 From this we can see that the vertical asymptotes occur at x = − 3 and x = 2. Next, we look at the behavior as x → ±∞ to find the horizontal asymptotes. (x + 2) (x + 2) 1/x + 2/x2 0+0 = lim = lim = =0 2 − x − 10 2 x→±∞ 3x x→±∞ 3 − 1/x − 10/x x→±∞ (3x + 5)(x − 2) 3−0−0 lim Therefore the horizontal asymptote is y = 0 4 (b) Graph f (x) making sure to label all asymptotes and important points. From above, we see that the there is a removable discontinuity at x = 3. So, we take the limit as x → 3 to locate the hole in the graph. lim x→ 3 5 5 (x + 2) = = (3x + 5)(x − 2) 14 · 1 14 Next we look at the be...
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