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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 Summer '08
 FBHinkelmann
 Calculus, Limits

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