2.6 - Review Key

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online